Answer:
Step-by-step explanation:
The proportion of students in a psychology experiment who could remember an eight-digit number correctly for t minutes was :
0.9 − 0.3 ln(t)
the proportion that remembered the number for 5 minutes, we would substitute t = 5 into expression, 0.9 − 0.3 ln(t). It becomes
0.9 − 0.3 ln5
= 0.9 - 0.3 × 1.60943791243
= 0.9 - 0.4828
= 0.4172
Approximately 0.4 to 1 decimal place
The population of a country has been decreasing for several decades. In 1990, the population was about 167 million people. In 2010, the population was about 163 million people. Determine the percent decrease in the country's population during this time.
The country's population decreased by about ___% during this time period (Round to one decimal place as needed)
The population of the country decreased by approximately 2.4% between 1990 and 2010, calculated using the percent decrease formula $(167 million - 163 million) / 167 million * 100$.
Explanation:To find the percent decrease in the country's population, we first need to determine the actual decrease in population. In this case, the population in 1990 was about 167 million and it decreased to 163 million in 2010, so the numeric decrease is 167 million - 163 million = 4 million.
Next, we compute the percent decrease based on the initial number from 1990. The formula for percent decrease is: (Decrease in Value / Original Value) * 100. Therefore, the percent decrease is (4 million / 167 million) * 100 = 2.4% (rounded to one decimal place).
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The decrease in the country's population from 1990 to 2010 is approximately 2.4%.
Explanation:To find the percentage decrease in the country's population, we first calculate the amount of decrease. This is done by subtracting the population in 2010 from the population in 1990, in other words, 167 million - 163 million, which equals 4 million. Next, we divide this number by the original population(the population in 1990) and times the result by 100%. Therefore, the percentage decrease in population is (4/167) * 100%, approximately equals 2.4% when rounded to one decimal place.
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A whole number has a 6 in the hundreds place, and all of its other digits are zero. If the 6 is moved from the hundreds place to the tens place and all other digits remain as zeros, which statement is true?
The new number is 10 times the original number.
The original number is 10 times the new number.
The new number is 100 times the original number.
The original number is 100 times the new number.
Answer:
The original number is 10 times the new number.
Step-by-step explanation:
An upright cylindrical tank with radius 8 m is being filled with water at a rate of 2 m3/min. How fast is the height of the water increasing? Part 1 of 3. If h is the water's height, the volume of the water is V = πr2h. We must find dV/dt. Differentiating both sides of the equation gives Dv/Dt= πr2 Dh/Dt Subsituting for r , this becomes Dv/Dt ____________ π Dh/Dt What goes in the blank ? Thanks !
Answer:
16 goes in the blank
Step-by-step explanation:
V(c) = 2*π*r*h
Differentiating boh sides
DV(c)/Dt = 2πr Dh/Dt now radius is 8 m
DV(c)/Dt = 8π Dh/Dt
That expression gives the relation of changes in V and h
DV(c)/Dt is the speed of growing of the volume
Dh/Dt is the speed of increase in height
so if the cylinder is filling at a rate of 2 m³/min the height will increase at a rate of 16π m/min
A company claims that the mean weight per apple they ship is 120 grams with a standard deviation of 12 grams. Data generated from a sample of 49 apples randomly selected from a shipment indicated a mean weight of 122.5 grams per apple. Is there sufficient evidence to reject the company’s claim? (useα = 0.05)
Answer:
There is no sufficient evidence to reject the company's claim at the significance level of 0.05
Step-by-step explanation:
Let [tex]\mu[/tex] be the true mean weight per apple the company ship. We want to test the next hypothesis
[tex]H_{0}:\mu=120[/tex] vs [tex]H_{1}:\mu\neq 120[/tex] (two-tailed test).
Because we have a large sample of size n = 49 apples randomly selected from a shipment, the test statistic is given by
[tex]Z=\frac{\bar{X}-120}{\sigma/\sqrt{n}}[/tex] which is normally distributed. The observed value is
[tex]z_{0}=\frac{122.5-120}{12/\sqrt{49}}=1.4583[/tex]. The rejection region for [tex]\alpha = 0.05[/tex] is given by RR = {z| z < -1.96 or z > 1.96} where the area below -1.96 and under the standard normal density is 0.025; and the area above 1.96 and under the standard normal density is 0.025 as well. Because the observed value 1.4583 does not fall inside the rejection region RR, we fail to reject the null hypothesis.
To determine whether there is sufficient evidence to reject the company's claim, we can perform a hypothesis test.
Explanation:To determine whether there is sufficient evidence to reject the company's claim, we can perform a hypothesis test. The null hypothesis, H0, is that the mean weight per apple is 120 grams, and the alternative hypothesis, Ha, is that the mean weight per apple is not 120 grams. With a sample of 49 apples, we can calculate the test statistic using the formula:
t = (sample mean - population mean) / (standard deviation / sqrt(sample size))
Once we have the test statistic, we can compare it to the critical value from the t-distribution to determine if we reject or fail to reject the null hypothesis. As per calculation, yhe area below -1.96 and under the standard normal density is 0.025; and the area above 1.96 and under the standard normal density is 0.025 as well.
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A ship leaves port at 12:30 pm and travels Upper N 55 degrees Upper E at the rate of 8 mph. Another ship leaves the same port at 1:30 pm and travels Upper N 30 degrees Upper W at the rate of 10 mph. How far apart in miles are the ships at 2:00 pm ?
Answer:
d=0.0167mille
Step-by-step explanation:
we must determine its position (X-Y).
ship 1
[tex]v_{1}=8\frac{m}{h}\\t_{1}=12.5 h\\t_{2}=14h\\t_{t2}=t_{2}-t_{1}=14-12.5=1.5h\\d_{s1}=v_{1}*t_{t1}=8\frac{m}{h}*1.5h=12m[/tex]
ship 2
[tex]v_{2}=10\frac{m}{h}\\t_{1}=12.5 h\\t_{2}=14h\\t_{t2}=t_{2}-t_{1}=14-13.5=0.5h\\d_{s2}=v_{1}*t_{t2}=10\frac{m}{h}*0.5h=5m[/tex]
component x-y
[tex]sin55=\frac{y_{1}}{h};y_{1}=12*sin55=9.82\\ cos55=\frac{x_{1} }{h};x_{1}=5*cos55=6,88[/tex]
[tex]sin150=\frac{y_{2}}{h};y_{2}=5*sin150=2.5\\cos150=\frac{x_{2} }{h};x_{2}=5*cos150=-4.33[/tex]
[tex]s_{1}=(6.88,9.82); s_{2}=(-4.33,2,5)[/tex]
we must find the distance S1-S2
[tex]d_{s1-s2}=\sqrt{(y_{2}-y_{1})^{2}+(x_{2}-x_{1})^{2}}=\sqrt{(2.5-9.8)^{2}+(-4.33-6.88)^{2}}\\=\sqrt{(24.5)^{2}+(-11.21)^{2}}=26.94m[/tex]
but the units are requested to be miles
d=26.94m*mille/1,609.34m=0.0167mille
[tex]97.28 \div 19[/tex]
An angle's initial ray points in the 3-o'clock direction and its terminal ray rotates CCW. Let θ represent the angle's varying measure (in radians).
If θ=0.5θ=0.5 what is the slope of the terminal ray?
If θ=1.78θ=1.78, what is the slope of the terminal ray?
Write an expression (in terms of θθ) that represents the varying slope of the terminal ray.
Answer:
0.546 , -4.71
Step-by-step explanation:
Given:
An angle's initial ray points in the 3-o'clock direction and its terminal ray rotates counter -clock wise.
Here, Slope = tan\theta
If θ = 0.5
Then, Slope = tan(θ) = tan(0.5) = 0.546
If θ = 1.78
Then, Slope = tan(θ) = tan(1.78) = - 4.71
The expression (in terms of θ) that represents the varying slope of the terminal ray.
Slope = m = tanθ, where θ is the varying angle
A) The slope of the terminal ray when θ = 0.5 radians is; 0.5463
B) The slope of the terminal ray when θ = 1.78 radians is; -4.71
C) The expression that will represent the varying slope of the terminal ray is;
tan (Δy/Δx)
We are given that θ represents the angle's varying measure (in radians).
Now, in mathematics, slope is simply the tangent of an angle. Thus;
A) At θ = 0.5 radians ,
Slope of terminal ray = tan θ
Slope = tan 0.5
Using radians calculator, tan 0.5 = 0.5463
Thus, slope = 0.5463
B) At θ = 1.78
Slope of terminal ray = tan θ
Slope = tan 1.78
Using radians calculator, tan 1.78 = -4.71
Thus, slope = -4.71
C) The expression that will represent the varying slope of the terminal ray is;
tan (Δy/Δx)
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8. Suppose you are testing H0 : p = 0.4 versus H1 : p < 0.4. From your data, you calculate your test statistic as z = +1.7. (a) Calculate the p-value for this scenario. (b) Using a significance level of 0.06, what decision should you make?
Answer:
(a) The p-value is 0.9554
(b) We cannot reject the null hypothesis at the significance level of 0.06
Step-by-step explanation:
We are dealing with a lower-tail alternative [tex]H_{1}: p < 0.4[/tex]. Because the test statistic is z = +1.7 which comes from a normal distribution, the p-value is the probability of getting a value as extreme as the already observed.
(a) P(Z < +1.7) = 0.9554
(b) The p-value is very large, and 0.9554 > 0.06, so, we cannot reject the null hypothesis at the significance level of 0.06
Which of the following is an example of a quantitative variable?
A) The color of an automobile
B) A person's zip code
C) A person's height, recorded in inches
D) Both B and C
Answer:
C) A person's height, recorded in inches
Step-by-step explanation:
Quantitative Variable:
A quantitative variable is a variable which can be measured and have a numeric outcome.That is the value of variable can be expressed with numbers.Foe example: age, length are examples of quantitative variables.A) The color of an automobile
The color of car is not a quantitative variable as its outcome cannot be measured and expressed in value. It is a categorical variable.
B) A person's zip code
Some variables like zip codes take numerical values. But they are not considered quantitative. They are considered as a categorical variable because average of zip codes have no significance.
C) A person's height, recorded in inches
Height is a qualitative variable because it can be measured and its value is expressed in numbers.
C) A person's height, recorded in inches
A variable can be defined as the characteristics of an object.
A quantitative variable is a variable that usually changes it values as a result of either counting or measuring. Examples are time, weight, height.
A qualitative variable is a variable which do not usually change as a result of counting or measurement. Such as the color of an object, zip code, phone number, license number.
Height is a quantitative variable.
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A student writes an incorrect step while checking if the sum of the measures of the two remote interior angles of triangle ABC below is equal to the measure of the exterior angle.
Step 1: m∠m + m∠n + m∠o = 180 degrees (sum of angles of a triangle)
Step 2: m∠p − m∠o = 90 degrees (alternate interior angles)
Step 3: Therefore, m∠m + m∠n + m∠o = m∠o + m∠p
Step 4: So, m∠m + m∠n = m∠p
In which step did the student first make a mistake and how can it be corrected?
1.) Step 1; it should be m∠m + m∠n + m∠o = 90 degrees (corresponding angles)
2.) Step 1; it should be m∠m + m∠n + m∠o = 90 degrees (adjacent angles)
3.) Step 2; it should be m∠o + m∠p = 180 degrees (alternate exterior angles)
4.) Step 2; it should be m∠o + m∠p = 180 degrees (supplementary angles)
Answer:
4.) Step 2; it should be m∠o + m∠p = 180 degrees (supplementary angles)
Step-by-step explanation:
o and p are supplementary angles, and therefore add up to 180 degrees.
A polymer is manufactured in a batch chemical process. Viscosity measurements are normally made on each batch, and long experience with the process has indicated that the variability in the process is fairly stable with o 20. Fifteen batch
viscosity measurements are given as follows:
724, 718, 776, 760, 745, 759, 795, 756, 742, 740, 761, 749, 739, 747, 742
A process change that involves switching the type of catalyst used in the process is made. Following the process change, eight batch viscosity measurements are taken:
735, 775, 729, 755, 783, 760, 738, 780
Assume that process variability is unaffected by the catalyst change. If the difference in mean batch viscosity is 10 or less, the manufacturer would like to detect it with a high probability.
(a) Formulate and test an appropriate hypothesis using a 0.10. What are your conclusions? Find the P-value.
(b) Find a 9096 confidence interval on the difference in mean batch viscosity resulting from the process change.
(c) Compare the results of parts (a) and (b) and discuss your findings.
Answer:
a) Null hypothesis:[tex]\mu_{s}-\mu_{n}\geq 10[/tex]
Alternative hypothesis:[tex]\mu_{s} - \mu_{n}<10[/tex]
[tex]p_v =P(Z<-1.904)=0.0284[/tex]
Comparing the p value with the significance level [tex]\alpha=0.1[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and the difference between the two groups is significantly lower than 10.
b) The 90% confidence interval would be given by [tex]-21.035 \leq \mu_1 -\mu_2 \leq 7.685[/tex]
c) If we analyze the interval obtained we see that the interval contains the value of -10 so then we agree with the result obtained from the hypothesis, that he alternative hypothesis is true.
Step-by-step explanation:
Data given and notation
[tex]\bar X_{s}=750.2[/tex] represent the mean for the sample standard process
[tex]\bar X_{n}=756.875[/tex] represent the mean for the sample with the new process
[tex]s_{s}=19.128[/tex] represent the sample standard deviation for the sample standard process
[tex]s_{n}=21.283[/tex] represent the sample standard deviation for the new process
[tex]\sigma_s=\sigma_n=\sigma=20[/tex] represent the population standard deviation for both samples.
[tex]n_{s}=15[/tex] sample size for the group Cincinnati
[tex]n_{n}=8[/tex] sample size for the group Pittsburgh
z would represent the statistic (variable of interest)
Concepts and formulas to use
(a) Formulate and test an appropriate hypothesis using a 0.10. What are your conclusions?
We need to conduct a hypothesis in order to check if the difference in mean batch viscosity is 10 or less system of hypothesis would be:
Null hypothesis:[tex]\mu_{s}-\mu_{n}\geq 10[/tex]
Alternative hypothesis:[tex]\mu_{s} - \mu_{n}<10[/tex]
We have the population standard deviation, so for this case is better apply a z test to compare means, and the statistic is given by:
[tex]z=\frac{(\bar X_{s}-\bar X_{n})-\Delta}{\sqrt{\frac{\sigma^2_{s}}{n_{s}}+\frac{\sigma^2_{n}}{n_{n}}}}[/tex] (1)
z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.
Calculate the statistic
With the info given we can replace in formula (1) like this:
[tex]z=\frac{(750.2-756.875)-10}{\sqrt{\frac{20^2}{15}+\frac{20^2}{8}}}}=-1.904[/tex]
Statistical decision
Since is a one tail left test the p value would be:
[tex]p_v =P(Z<-1.904)=0.0284[/tex]
Comparing the p value with the significance level [tex]\alpha=0.1[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and the difference between the two groups is significantly lower than 10.
(b) Find a 90% confidence interval on the difference in mean batch viscosity resulting from the process change.
The confidence interval for the difference of means is given by the following formula:
[tex](\bar X_s -\bar X_n) \pm z_{\alpha/2}\sqrt{\sigma^2(\frac{1}{n_s}+\frac{1}{n_s})}[/tex] (1)
The point of estimate for [tex]\mu_1 -\mu_2[/tex] is just given by:
[tex]\bar X_s -\bar X_n =750.2-756.875=-6.675[/tex]
Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that [tex]z_{\alpha/2}=1.64[/tex]
The standard error is given by the following formula:
[tex]SE=\sqrt{\sigma^2(\frac{1}{n_s}+\frac{1}{n_n})}[/tex]
And replacing we have:
[tex]SE=\sqrt{20^2(\frac{1}{15}+\frac{1}{8})}=8.756[/tex]
Confidence interval
Now we have everything in order to replace into formula (1):
[tex]-6.675-1.64\sqrt{20^2(\frac{1}{15}+\frac{1}{8})}=-21.035[/tex]
[tex]-6.675+1.64\sqrt{20^2(\frac{1}{15}+\frac{1}{8})}=7.685[/tex]
So on this case the 90% confidence interval would be given by [tex]-21.035 \leq \mu_1 -\mu_2 \leq 7.685[/tex]
(c) Compare the results of parts (a) and (b) and discuss your findings.
If we analyze the interval obtained we see that the interval contains the value of -10 so then we agree with the result obtained from the hypothesis, that he alternative hypothesis is true.
Final answer:
The viscosity analysis involves a hypothesis test to check for significant changes in mean batch viscosity and constructing a confidence interval to estimate the range of this difference following a process change in the manufacturing of a polymer.
Explanation:
The question involves conducting a hypothesis test and constructing a confidence interval to analyze the change in mean batch viscosity of a polymer before and after a process change. A hypothesis test will be used to determine if the mean batch viscosity has changed significantly, and the confidence interval will provide a range in which the true difference in means likely falls.
Hypothesis Test:
Calculate the mean of both sets of viscosity measurements.
Set up the null hypothesis (H0) and the alternative hypothesis (H1): H0: μ1 - μ2 <= 10 (no significant change), H1: μ1 - μ2 > 10 (significant change).
Determine the test statistic using a t-test for two independent samples.
Calculate the p-value associated with the test statistic.
Compare the p-value with the significance level (α = 0.10). If p <= α, reject H0; otherwise, fail to reject H0.
Calculate the standard deviation of both sets of measurements.
Determine the standard error of the difference in means.
Find the appropriate t-distribution critical value based on the given confidence level (90%) and degrees of freedom.
Calculate the 90% confidence interval using the difference in means ± the margin of error.
Compare the results of the hypothesis test and the confidence interval. If the hypothesis test leads to rejection of the null hypothesis while the confidence interval for the difference in means includes values greater than 10, there is evidence suggesting a significant change in mean batch viscosity. However, if the hypothesis test does not lead to rejection of H0 and the confidence interval includes 10, it suggests that the process change did not have a significant effect on mean batch viscosity.
Suppose we take the interval [−5,7] and divide it into 4 equal subintervals. Find the width Δn of each subinterval. Δn = If we name the endpoints of the subintervals x0, x1, x2, x3 and x4 , with x0 on the left and x4 on the right, find the values of these endpoints and list them in ascending order. (Enter your answers in a comma-separated list.)
Answer:
[tex]x_0=-5\\x_1=-2\\x_2=1\\x_3=4\\x_4=7[/tex]
Δn=3
Step-by-step explanation:
Remember, if we need to divide the interval (a,b) in n equal subinterval, then we need divide the distance (d) between the endpoints of the interval and divide it by n. Then the width Δn of each subinterval is d/n.
We have the interval [-5,7]. The distance between the endpoints of the interval is
[tex]d=7-(-5)=12[/tex].
Now, we divide d by 4 and obtain [tex]\frac{d}{4}=\frac{12}{4}=3[/tex]
Then, Δn=3.
Now, to find the endpoints of each sub-interval, we add 3 from the left end of the interval.
[tex]-5=x_0\\x_0+3=-5+3=-2=x_1\\x_1+3=1=x_2\\x_2+3=4=x_3\\x_3+3=7=x_4[/tex]
So,
[tex]x_0=-5\\x_1=-2\\x_2=1\\x_3=4\\x_4=7[/tex]
The width of each subinterval is 3. The endpoints in ascending order are -5, -2, 1, 4, 7.
Explanation:To find the width of each subinterval, we need to divide the length of the interval by the number of subintervals. In this case, the interval is [-5,7] and we need to divide it into 4 equal subintervals. The length of the interval is 7 - (-5) = 12. So the width of each subinterval is 12/4 = 3.
Next, we can find the endpoints of the subintervals. Since we have 4 subintervals, we need 5 endpoints. The first endpoint is the left endpoint of the interval, which is -5. Then we add the width of each subinterval to find the next endpoints: -5 + 3 = -2, -2 + 3 = 1, 1 + 3 = 4, and finally 4 + 3 = 7, which is the right endpoint of the interval. Therefore, the endpoints in ascending order are -5, -2, 1, 4, 7.
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A bicyclist is training for a race on a hilly path. Their bike keeps track of their speed at any time, but not the distance traveled. Their speed traveling up a hill is 3mph, 8mph when traveling down a hill, and 5mph when traveling along a flat portion. Part A. Construct linear models that describe their distance, D in miles, on a particular portion of the path in terms of the time, t in hours, spent on that part of the path.
Answer:
[tex]D_{up}(t)=3t[/tex]
[tex]D_{down}(t)=8t[/tex]
[tex]D_{flat}(t)=5t[/tex]
Explanation:
To construct a linear model of their distance, in miles, as a function of time in hours, since we were given their traveling speed in miles per hour, simply multiply the traveling speed in mph by the time, t, in hours.
When traveling up a hill:
[tex]D_{up}(t)=3t[/tex]
When traveling down a hill:
[tex]D_{down}(t)=8t[/tex]
When traveling along a flat portion:
[tex]D_{flat}(t)=5t[/tex]
Final answer:
To construct linear models that describe the distance traveled on different portions of the path in terms of time, we can use the formula for distance: D = V * T, where D is the distance, V is the velocity, and T is the time.
Explanation:
To construct linear models that describe the distance traveled on different portions of the path in terms of time, we can use the formula for distance:
D = V * T
where D is the distance, V is the velocity, and T is the time.
For the uphill portion, the speed is 3 mph, so the linear model would be:
D = 3T
For the downhill portion, the speed is 8 mph, so the linear model would be:
D = 8T
For the flat portion, the speed is 5 mph, so the linear model would be:
D = 5T
In 2016, among the top 100 grossing movies in a particular country, 16 were rated PG-13 and earned over $100 million. The number of movies that were rated PG-13 that earned less than $100 million was 29. The number of movies that were not rated PG-13 that earned less than $100 million was 39. What was the number of movies that earned over $100 million and that were not rated PG-13?
Answer:
16
Step-by-step explanation:
Let,
P = movie rated PG-13,
P' = movie does not rated PG-13,
E = movie earned over $100 million,
E' = movie earned less than $100 million,
According to the question,
[tex]n(P\cap E)=16[/tex]
[tex]n(P\cap E')=29[/tex]
[tex]n(P'\cap E')=39[/tex]
Since,
[tex]n(P\cap E)+n(P\cap E')+n(P'\cap E')+n(P'\cap E)[/tex] = total movies
[tex]16 + 29 + 39 + n(P'\cap E) = 100[/tex]
[tex]n(P'\cap E) = 100 - 84 = 16[/tex]
Hence,
The number of movies that earned over $100 million and that were not rated PG-13 would be 16.
To find the number of movies that earned over $100 million but were not rated PG-13, subtract the number of PG-13 rated movies that earned over $100 million from the total number of movies that earned over $100 million.
Explanation:To find the number of movies that earned over $100 million but were not rated PG-13, we need to subtract the number of PG-13 rated movies that earned over $100 million from the total number of movies that earned over $100 million.
The total number of movies that earned over $100 million is given by the sum of the number of PG-13 rated movies that earned over $100 million (16) and the number of movies that were not rated PG-13 that earned over $100 million.From the information given, we know that the number of movies that were rated PG-13 that earned less than $100 million is 29, and the number of movies that were not rated PG-13 that earned less than $100 million is 39.Therefore, the number of movies that were not rated PG-13 that earned over $100 million can be found by subtracting the number of PG-13 rated movies that earned over $100 million from the total number of movies that earned over $100 million: Total movies earning over $100 million - PG-13 rated movies earning over $100 million = Movies not rated PG-13 earning over $100 million.Substituting the given values, we have:
Total movies earning over $100 million - 16 = Movies not rated PG-13 earning over $100 million.
Therefore, the number of movies that earned over $100 million and were not rated PG-13 is 100 - 16 = 84.
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When the student summarizes the data, she finds that 42 of the 50 business students and 38 of the 70 nursing students admitted to cheating in their courses. True or false? The counts suggest that the normal model is a good fit for the sampling distribution of sample differences.
Answer:
True
Step-by-step explanation:
The answer is "True" because the data set was picked at random.
A geologist examines 6 seawater samples for lead concentration. The mean lead concentration for the sample data is 0.903 cc/cubic meter with a standard deviation of 0.0566 . Determine the 95% confidence interval for the population mean lead concentration. Assume the population is approximately normal.
Step 2 of 2 :
Construct the 95% confidence interval. Round your answer to three decimal places.
Answer:
Step-by-step explanation:
We want to determine a 95% confidence interval for the mean lead concentration of sea water samples
Number of samples. n = 6
Mean, u = 0.903 cc/cubic meter
Standard deviation, s = 0.0566
For a confidence level of 95%, the corresponding z value is 1.96. This is determined from the normal distribution table.
We will apply the formula
Confidence interval
= mean +/- z ×standard deviation/√n
It becomes
0.903 +/- 1.96 × 0.0566/√6
= 0.903 +/- 1.96 × 0.0566/2.44948974278
= 0.903 +/- 0.045
The lower end of the confidence interval is 0.903 - 0.045 =0.858
The upper end of the confidence interval is 0.903 + 0.045 =0.948
Therefore, with 95% confidence interval, the mean lead concentration of the sea water is between 0.858 cc/cubic meter and 0.948 cc/cubic meter
What is the relationship between probability of type 1 error and probability of type 2 error?
A) As α increases, β decreases
B) As α decreases, β decreases
C) The relationship depends on the context of the hypothesis test
D) Both α and β are independent of one anothe
Answer:
A) As α increases, β decreases
Step-by-step explanation:
I am assuming that the probabilities of both kind of error are positive, otherwise there is no reason to make an hypothesis test.
Both events arent possible simultaneously, so they cant be independent because the probability of both kind of errors is greater than 0.
If the p-value increases, then by definition, the probability of type 1 error increases, in this case, you are more likely to reject the null hypothesis, therefore you are less likely to make the mistake of not rejecting the null hypothesis when you have to. Thus, the probability of type 2 error decreases.
On the other way around, if you are less likely to reject the null hypothesis (hence, the probability of type 1 error decreases), then you will be more likely to not reject the null hypothesis even on weird scenarios, therefore the probability of type 2 error increases.
There is no formula to determine the relatioship between both errors, but the errors are related: when the probability of type 1 error increases, then the probability of type 2 error decreases. The correct answer is A)
According to a study conducted in one city, 35% of adults in the city have credit card debts more than $2.000. A simple random sample of n=250 adults is obtained from the city. Describe the sampling distribution of P^, the sample proportion of adults who have credit card debts of more than $2000.(Round to three decimal places when necessary.)Select from one of the 4 answers belowA. approximately- normal; \mu p=0.35, \sigma p=0.030B. approximately- normal; \mu p=0.35, \sigma p=0.001C. exactly- normal; \mu p=0.35, \sigma p=0.030D.Binomial; \mu p=87.5, \sigma p=7.542
Final Answer:
A. approximately-normal; μp = 0.35, σp = 0.030
1. Identify conditions for a normal sampling distribution:
Random sampling: The problem states it's a simple random sample.
Large sample size: n = 250 is greater than 10% of the population (assumed to be large), satisfying the condition.
Success-failure condition: np = 250 * 0.35 = 87.5 and n(1-p) = 250 * 0.65 = 162.5 are both greater than 10, meeting the condition.
2. Calculate the mean and standard deviation of the sampling distribution:
Mean (μp): μp = p = 0.35 (equal to the population proportion)
Standard deviation (σp): σp = sqrt(p(1-p)/n) = sqrt(0.35*0.65/250) ≈ 0.030
Final Answer:
A. approximately-normal; μp = 0.35, σp = 0.030
Explanation:
The sampling distribution of P^ is approximately normal due to the satisfaction of the conditions mentioned above.The mean of the sampling distribution is equal to the population proportion (0.35).The standard deviation of the sampling distribution is calculated as 0.030.a. Suppose that a single card is selected from a standard 52-card deck. What is the probability that the card drawn is a spade?
b. Now suppose that a single card is drawn from a standard 52-card deck, but it is told that the card is black.What is the probability that the card drawn is a spade?
The probability of drawing a spade from a 52-card deck is 0.25. If it's known that the card is black, the probability that it's a spade increases to 0.5.
Explanation:The subject of this question is the calculation of probability in a card game scenario. When a single card is selected from a standard 52-card deck, the probability that the card drawn is a spade is calculated by dividing the total number of spades in the deck (13) by the total number of cards in the deck (52). This gives a probability of 13/52 or 0.25.
However, if a single card is drawn from a standard 52-card deck, and it is known that the card is black, then the total number of possible outcomes is reduced to 26 (the number of black cards in the deck). In this scenario, the probability that the card drawn is a spade becomes 13/26 or 0.5.
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The probability of drawing a spade from a 52-card deck is 1/4. If we already know the card is black, the probability increases to 1/2.
Explanation:This question relates to the field of probability. To address part a, in a standard 52-card deck, there are 13 spades. We therefore have a 13 out of 52 chance to draw a spade, simplifying to a probability of 1/4.
For part b, being told that the card drawn is black means our sample space becomes 26 (13 spades and 13 clubs). Since there are still 13 spades, the probability of drawing a spade given that the card is black is 13 out of 26, or 1/2.
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Suppose are running a study/poll about the proportion of voters who prefer Candidate A. You randomly sample 86 people and find that 59 of them match the condition you are testing.
Suppose you are have the following null and alternative hypotheses for a test you are running:
H0:p=0.68H0:p=0.68
Ha:p<0.68Ha:p<0.68
Calculate the test statistic, rounded to 3 decimal places
Answer:
[tex]z=\frac{0.686 -0.68}{\sqrt{\frac{0.68(1-0.68)}{86}}}=0.119[/tex]
[tex]p_v =P(z<0.119)=0.547[/tex]
The p value obtained was a very high value and using the significance level given for example [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion interest is not significantly lower than 0.68.
Step-by-step explanation:
1) Data given and notation
n=86 represent the random sample taken
X=59 represent the adults that match the condition you are testing
[tex]\hat p=\frac{59}{86}=0.686[/tex] estimated proportion of adults that match the condition you are testing
[tex]p_o=0.68[/tex] is the value that we want to test
[tex]\alpha[/tex] represent the significance level
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value (variable of interest)
2) Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that the population proportion is less than 0.68.:
Null hypothesis:[tex]p=0.68[/tex]
Alternative hypothesis:[tex]p < 0.68[/tex]
When we conduct a proportion test we need to use the z statistic, and the is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].
3) Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.686 -0.68}{\sqrt{\frac{0.68(1-0.68)}{86}}}=0.119[/tex]
4) Statistical decision
It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.
The significance level provided is not given [tex]\alpha[/tex]. The next step would be calculate the p value for this test.
Since is a left tailed test the p value would be:
[tex]p_v =P(z<0.119)=0.547[/tex]
So the p value obtained was a very high value and using the significance level given for example [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion interest is not significantly lower than 0.68.
Monochromatic light from a helium-neon laser of wavelength of 632.8 nm is incident normally on a diffraction grating containing 6000 lines/cm. Find the angles at which one would observe the first order maximum, the second-order maximum, and so forth.
To determine the angles of diffraction maxima for a helium-neon laser light incident on a diffraction grating, use the formula d sin(θ) = m λ with d the grating spacing and λ the wavelength. Calculate for each order by substituting m=1, 2, etc., and solve for θ using inverse sine function.
Explanation:To find the angles at which one would observe the first order maximum, second-order maximum, and so forth for monochromatic light incident on a diffraction grating, we use the diffraction grating equation: d × sin(θ) = m × λ, where d is the grating spacing, λ is the wavelength of light, m is the order of maximum, and θ is the diffraction angle.
For a diffraction grating with 6000 lines/cm, the spacing d is 1/6000 cm, or 1.67 x 10^-4 cm (since 1 cm = 10^-2 m, d = 1.67 x 10^-6 m). Given the wavelength λ = 632.8 nm or 6.328 x 10^-7 m, we can substitute these values into the equation to solve for θ for any order m.
For the first order maximum (m=1), solve d × sin(θ) = 1 × λ.For the second order maximum (m=2), solve d × sin(θ) = 2 × λ.And so forth for higher orders.To find the actual angles, use the inverse sine function (arcsin), keeping in mind the units.
Based on a poll, 67% of Internet users are more careful about personal information when using a public Wi-Fi hotspot. What is the probability that among four randomly selected Internet users, at least one is more careful about personal information when using a public Wi-Fi hotspot? How is the result affected by the additional information that the survey subjects volunteered to respond?
Answer:
The required probability is 0.988.
Step-by-step explanation:
Consider the provided information.
Based on a poll, 67% of Internet users are more careful about personal information when using a public Wi-Fi hotspot.
That means the probability of more careful is 0.67
The probability of not careful is: 1-0.67 = 0.33
We have selected four random Internet users. we need to find the probability that at least one is more careful about personal information.
P(At least one careful) = 1 - P(None of them careful)
P(At least one careful) = 1 - (0.33×0.33×0.33×0.33)
P(At least one careful) = 1 - 0.012
P(At least one careful) = 0.988
Hence, the required probability is 0.988.
The result may be higher because of the convenience bias in retrieving the sample. Because the survey subjects volunteered to respond not random.
The probability that at least one out of four randomly selected Internet users is more careful about personal information when using a public Wi-Fi hotspot is approximately 98.81%, assuming that each event is independent and the probability of an individual being careful is 67%. However, voluntary responses to the survey might introduce a non-response bias, affecting the accuracy of this probability.
Explanation:The question asks for the probability that among four randomly selected Internet users, at least one is more careful about personal information when using a public Wi-Fi hotspot, given that 67% of Internet users are like this. To find the probability of 'at least one', it is easier to calculate the complement—that is, the probability that none of the four users are careful—and subtract it from 1 (the total probability of any outcome).
The probability that a randomly selected Internet user is not more careful is 1 - 0.67 = 0.33. Since we are considering four independent events, we raise the single-event probability to the fourth power:
(0.33)^4 = 0.33 * 0.33 * 0.33 * 0.33
The calculation results in approximately 0.0119. Now, subtract this from 1 to get the probability of at least one person being more careful:
1 - 0.0119 = 0.9881
Therefore, the probability that at least one out of four randomly selected Internet users is more careful about personal information when using a public Wi-Fi hotspot is about 98.81%.
However, the accuracy of the result could be influenced by the fact that the survey subjects volunteered to respond, which can result in a non-response bias. Volunteers might have different behaviors or opinions compared to the general Internet user population, potentially skewing the results of the survey and, consequently, the estimated probability.
About Here are two relations defined on the set {a, b, c, d}: S = { (a, b), (a, c), (c, d), (c, a) } R = { (b, c), (c, b), (a, d), (d, b) } Write each relation as a set of ordered pairs. (a) S ο R (b) R ο S (c) S ο S
Answer with Step-by-step explanation:
We are given that a set {a,b,c,d}
S={(a,b),(a,c),(c,d),(c,a)}
R={(b,c),(c,b),(a,d),(d,b)]
Composition of relation:Let R and S are two relations on the given set
If ordered pair (a,b) belongs to relation R and (b,c) belongs to S .
Then, SoR={(a,c)}
By using this rule
SoR={(b,d),(b,a)}[/tex]
Because [tex](b,c)\in R[/tex] and [tex](c,d)\in S[/tex].Thus, [tex](b,d)\in SoR[/tex]
[tex](b,c))\in R[/tex] and [tex](c,a)\in S[/tex].Thus, [tex](b,a)\in SoR[/tex]
b.RoS={(a,c),(a,b),(c,b),(c,d)}
Because
[tex](a,b)\in S,(b,c)\in R[/tex] .Therefore, the ordered pair [tex](a,c)\in[/tex] RoS
[tex](a,c)\in S,(c,b)\in R[/tex] .Thus, [tex](a,b)\in RoS[/tex]
[tex](c,d)\in S,(d,b)\in R[/tex].Thus, [tex](c,b)\in RoS[/tex]
[tex](c,a)\in S,(a,d)\in R[/tex].Thus,[tex](c,d)\in RoS[/tex]
c.SoS={(a,d),(a,a),(c,c),(c,b)}
Because
[tex](a,c)\;and\; (c,d)\in S[/tex].Thus, [tex](a,d)\in SoS[/tex]
[tex](c,a),(a,b)\in S[/tex].Thus,[tex](c,b)\in SoS[/tex]
[tex](a,c)\in S[/tex] and [tex](c,a)\in S[/tex].Thus,[tex](a,a)\in SoS[/tex]
[tex](c,a)\in S[/tex] and [tex](a,c)\in S[/tex].Thus ,[tex](c,c)\in SoS[/tex]
Seven thousand lottery tickets are sold for $5 each. One ticket will win $2,000, two tickets will win $750 each, and five tickets will win $100 each. Let X denote the net gain from the purchase of a randomly selected ticket. a. Construct the probability distribution of X. b. Compute the expected value E(X) of X. Interpret its meaning. c. Compute the standard deviation σ of X.
Answer:
a) The distribution for the random variable X is given by:
X | -5 | 95 | 745 | 1995 |
P(X) | 6992/7000 | 5/7000 | 2/7000 | 1/7000 |
b) E(X)=-4.43. That means if we buy an individual ticket by $5 on this lottery the expected value of loss if $4.43.
c) [tex]Sd(X)=\sqrt{Var(X)}=\sqrt{738.947}=27.184[/tex]
Step-by-step explanation:
In statistics and probability analysis, the expected value "is calculated by multiplying each of the possible outcomes by the likelihood each outcome will occur and then summing all of those values".
The variance of a random variable Var(X) is the expected value of the squared deviation from the mean of X, E(X).
And the standard deviation of a random variable X is just the square root of the variance.
Part a
The info given is:
N=7000 represent the number of tickets sold
$5 is the price for any ticket
Number of tickets with a prize of $2000 =1
Number of tickets with a prize of $750=2
Number of tickets with a prize of $100=5
Let X represent the random variable net gain when we buy an individual ticket. The possible values that X can assume are:
___________________________
Ticket price Prize Net gain (X)
___________________________
5 2000 1995
5 750 745
5 100 95
5 0 -5
___________________________
Now we can find the probability for each value of X
P(X=1995)=1/7000, since we ave just one prize of $2000
P(X=745)=2/7000, since we have two prizes of $750
P(X=95)=5/7000, since we have 5 prizes of $100
P(X=-5)=6992/7000. since we have 6992 prizes of $0.
So then the random variable is given by this table
X | -5 | 95 | 745 | 1995 |
P(X) | 6992/7000 | 5/7000 | 2/7000 | 1/7000 |
Part b
In order to calculate the expected value we can use the following formula:
[tex]E(X)=\sum_{i=1}^n X_i P(X_i)[/tex]
And if we use the values obtained we got:
[tex]E(X)=(-5)*(\frac{6992}{7000})+(95)(\frac{5}{7000})+(745)(\frac{2}{7000})+(1995)(\frac{1}{7000})=\frac{-31000}{7000}=-4.43[/tex]
That means if we buy an individual ticket by $5 on this lottery the expected value of loss if $4.43.
Part c
In order to find the standard deviation we need to find first the second moment, given by :
[tex]E(X^2)=\sum_{i=1}^n X^2_i P(X_i)[/tex]
And using the formula we got:
[tex]E(X^2)=(25)*(\frac{6992}{7000})+(9025)(\frac{5}{7000})+(555025)(\frac{2}{7000})+(3980025)(\frac{1}{7000})=\frac{5310000}{7000}=758.571[/tex]
Then we can find the variance with the following formula:
[tex]Var(X)=E(X^2)-[E(X)]^2 =758.571-(-4.43)^2 =738.947[/tex]
And then the standard deviation would be given by:
[tex]Sd(X)=\sqrt{Var(X)}=\sqrt{738.947}=27.184[/tex]
Fill in the blank. (Note: The notation x → c+ indicates that x approaches c from the right and x → c− indicates that x approaches c from the left. Enter your answer in radians.) As x → 1−, the value of arcsin(x) → _____ .
Answer:
As x approaches 1 from the left, the value of [tex]arcsin(x)[/tex] is [tex]\frac{\pi }{2}[/tex]
Step-by-step explanation:
To find the value of [tex]arcsin(x)[/tex] when x approaches 1 from the left, you can use the graph of the function [tex]arcsin(x)[/tex]
Examine what happens as x approaches from the left.
As x approaches 1 from the left, the function seems to be approaching [tex]\frac{\pi }{2}[/tex]
Therefore, as x approaches 1 from the left, the value of [tex]arcsin(x)[/tex] is [tex]\frac{\pi }{2}[/tex]
As x approaches 1 from the left, the value of arcsin(x) approaches π/2 radians.
Explanation:The expression x → 1− refers to the limit as x approaches 1 from the left. For the function presented, arcsin(x), we need to consider the value that this function approaches as x gets infinitely close to 1, but is still less than 1. This is asking what angle in radians has a sine of 1, or almost 1 from the left side. The sine of an angle is 1 when that angle is π/2 (or 90° in degree measure), hence as x → 1−, the value of arcsin(x) approaches π/2.
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Babcock and Marks (2010) reviewed survey data from 2003–2005 and obtained an average of μ = 14 hours per week spent studying by full-time students at four-year colleges in the United States. To determine whether this average has changed in the past 10 years, a researcher selected a sample of n = 64 of today’s college students and obtained an average of M = 12.5 hours. If the standard deviation for the distribution is σ = 4.8 hours per week, does this sample indicate a significant change in the number of hours spent studying?
Answer:
We do not have enough evidence to accept H₀
Step-by-step explanation:
Normal Distribution
size sample = n = 64 (very small sample for evaluating population of 5 years
Standard deviation 4,8
1.- Test hypothesis
H₀ null hypothesis ⇒ μ₀ = 14 and
Hₐ alternative hypothesis ⇒ μ₀ ≠ 14
2.- z(c) we assume α = 0,05 as we are dealing with a two test tail we should consider α/2 = 0.025.
From z table we the z(c) value
z(c) = 1.96 and of course by symmetry z(c) = -1.96
3.- We proceed to compute z(s)
z(s) = [ ( μ - μ₀ ) /( σ/√n) ] ⇒ z(s) = - (1.5)*√64/4.8
z(s) = - 2.5
We compare z(s) and z(c)
z(s) < z(c) -2.5 < -1.96 meaning z(s) is in the rejection zone
we reject H₀ .
From the start we indicate sample size as to small for the experiment nonetheless we found that we dont have enough evidence to accept H₀
A civil engineer is analyzing the compressive strength of concrete. Compressive strength is approximately normally distributed with variance \sigma ^{2} = 1000(psi) 2. A random sample of 12 specimens has a mean compressive strength of Mean \mu= 3250psi.
a.) Construct a 95% two-sided confidence interval on mean compressive strength.
b.) Construct a 99% two-sided confidence interval on the mean compressive strength. Compare the width of this confidence interval with the width of the one found in a)
Answer:
solution is in the image below
Step-by-step explanation:
Final answer:
To construct confidence intervals for the mean compressive strength of concrete with known variance, formulas involving the z-score are employed. The 95% confidence interval is narrower than the 99% interval, illustrating that higher confidence requires a wider interval to encapsulate the true mean with more assurance.
Explanation:
To construct a confidence interval for the mean compressive strength of concrete, we utilize the formula for the confidence interval of the mean when the population variance (σ2) is known. Given that the variance is 1000 psi2 and the mean (μ) is 3250 psi for a sample of n=12 specimens, the z-score corresponding to a 95% confidence level is approximately 1.96, and for a 99% confidence level, the z-score is approximately 2.576.
95% Confidence Interval:
CI = μ ± z(σ/√n)
= 3250 ± 1.96(/1000/12)
= 3250 ± (1.96)(28.8675)
= 3250 ± 56.61
= (3193.39, 3306.61) psi
99% Confidence Interval:
CI = μ ± z(σ/√n)
= 3250 ± 2.576(/1000/12)
= 3250 ± (2.576)(28.8675)
= 3250 ± 74.36
= (3175.64, 3324.36) psi
Comparing the widths, the 99% confidence interval is wider than the 95% confidence interval, which is a reflection of increased certainty (or confidence level) requiring a wider interval.
Trials in an experiment with a polygraph include 96 results that include 22 cases of wrong results and 74 cases of correct results. Use a 0.01 significance level to test the claim that such polygraph results are correct less than 80% of the time Based on the results should polygraph test results be prohibited as evidence in trials? Identify the null hypothesis, alternative hypothesis, test statistics, p value, conclusion about null hypotheses and final conclusion that address the original claim. Use the p value methos. Use the normal distribution as an approximation of the binomial distribution. A. I dentify the Null and alternative hypothesis B. The test staistic is z= (round to two decimals) C. The P value is=( rounds to four decimals) D. Identify the conclusion about the null hypotheses and the final conclusion that address the original claim. Choose (fail to reject, reject) h0. There choose(is, is not) sufficient evidence to support the claim that the polygraph result are correct less than 80% of time. answer H0: P ≥ 0.80 Ha: P < 0.80 Estimated p = 74 / 98 = 0.7551 Variance of proportion = p*(1-p)/n = 0.8(0.2)/98 =0.0016327 S.D. of p is sqrt[0.001633] = 0.0404 z = ( 0.7551 - 0.8 ) / 0.0404 = -1.1112 P-value = P( z < -1.1112) = 0.1335 Since the p-value is greater than 0.05, we do not reject the null hypothesis. Based on the results there is no evidence that polygraph test results should be prohibited as evidence in trials.
Final answer:
Null Hypothesis (H0): P >= 0.80, Alternative Hypothesis (Ha): P < 0.80, Test Statistic (z): -1.1112, P-value: 0.1335, Conclusion: Fail to reject H0.
Explanation:
Null Hypothesis (H0): P ≥ 0.80
Alternative Hypothesis (Ha): P < 0.80
Test Statistic (z): -1.1112
P-value: 0.1335
Conclusion: Fail to reject H0. There is not sufficient evidence to support the claim that polygraph test results are correct less than 80% of the time.
Pam is an employee at a jewelry kiosk in a mall. If Pam works hard, there is a 75% probability that jewelry profits will equal $400 a day and a 25% probability that jewelry profits will equal $200 a day. If Pam shirks, there is a 25% probability that jewelry profits will equal $400 a day and a 75% probability that jewelry profits will equal $200 a day. Suppose Pam is paid 50% of the daily jewelry profits. What is Pam’s expected gain from working hard?
Answer:
175$
Step-by-step explanation:
Given that Pam is an employee at a jewelry kiosk in a mall. If Pam works hard, there is a 75% probability that jewelry profits will equal $400 a day and a 25% probability that jewelry profits will equal $200 a day. If Pam shirks, there is a 25% probability that jewelry profits will equal $400 a day and a 75% probability that jewelry profits will equal $200 a day.
Since Pam works hard
expected value of profit for the kiosk = [tex]400(0.75)+200(0.25)\\=350[/tex]
From this value, 50% would be given to Pam
Hence expected gain for Pam for working hard
= [tex]0.5(350)\\=175[/tex]
175 $
A set of 20 cards consists of 12 red cards and 8 black cards. The cards are shuffled thoroughly and you choose one at random, observe its color, and replace it in the set. The cards are thoroughly reshuffled, and you again choose a card at random, observe its color, and replace it in the set. This is done a total of six times. Let X be the number of red cards observed in these six trials. The variance of X isA. 6.B. 3.60.C. 2.4.D. 1.44.
Answer:
the answer D=1.44
Step-by-step explanation:
if the X= number of red cards observed in 6 trials , since each card observation is independent from the others and the sampling process is done with replacement ( the card is observed, then returned and reshuffled) , X follows an binomial distribution.
X(x)= n!/((n-x)!*x!) *p^x *(1-p)^(n-x)
where n = number of trials = 6 , x= number of red cards observed , p= probability of obtaining a red card in one try
the probability of obtaining the card in one try is
p = number of red cards / total number of cards = 12/ (12+8) = 0.6
since we know that X has a binomial distribution, the variance of this kind of distribution is
variance = σ² = n * p * (1-p)
therefore the variance of X is
variance = σ² = n * p * (1-p) = 6 * 0.6 * (1-0.6) = 1.44