Answers
Answer:
mean = 70000
SD = 15239
X= 95000
Z = (X-mean)/ SD
= (95000-70000)/15239
= 1.64
Now from Z-Table
% of employees = 100(1-.9495) = 5.02%
Related Questions
Scores on a recent national statistics exam were normallydistributed with a mean of 80 and a standard deviation of 6.a) What is the probability that a randomly selected examwill have a score of at least 71?b) What percentage of exams will have scores between 89and 92?c) If the top 2.5% of test scores receive merit awards,what is the lowest score eligible for an award?d) If there were 334 exams with scores of at least 89,how many students took the exam?
Answers
A study was conducted to determine if the salaries of elementary school teachers from two neighboring districts were equal. A sample of 15 teachers from each district was randomly selected. Test the claim that the salaries from both districts are equal. Use significance 0.05 and assume that the data are distributed normally. n x¯ s District 1: 15 $28,900 $2,300 District 2: 15 $30,300 $2,100 3A. What is the appropriate set of hypotheses (H0, H1)?
• µ1 – µ2 = 0, µ1 – µ2 ≠ 0
• µ1 – µ2 ≠ 0, µ1 – µ2 = 0
• µ1 – µ2 = 0, µ1 – µ2 < 0
• µ1 – µ2 = 0, µ1 – µ2 > 0 3B.
What is the correct confidence interval?
• (–2475, –325)
• (–2100, –700)
• (–2975, 175)
• (–3125, 325) 3C.
Are the salaries of the teachers from the two districts different?
• Yes.
• No.
• Sometimes.
• Depends on distribution.
Answers
Answer:
H₀: μ₁ - μ₂ = 0
H₁: μ₁ - μ₂ ≠ 0
(–2975, 175)
Do not reject H₀
Step-by-step explanation:
Hello!
The objective of this experiment is to test if the salaries of elementary school teachers are equal in two districts. You can test this trough the population means of the salaries of the teachers if they are either equal or different or directly test if the difference between the salaries of the two districts is cero or not, symbolically: μ₁ - μ₂ = 0
Remember, in the null hypothesis is usually stated the known information, is the "no change" premise and always carries the = symbol.
The hypothesis is:
H₀: μ₁ - μ₂ = 0
H₁: μ₁ - μ₂ ≠ 0
You have two normally distributed variables and you are studying the difference of the means. You can use a pooled Z to make the interval, the formula is:
X[bar]₁-X[bar]₂ ± [tex]Z_{1-\alpha /2}[/tex]*(√(σ₁²/n₁+σ₂²/n₂)
Since the test is two-tailed and at a signification level of 5% I've made the interval at the complementary confidence level of 95% so that I can use it to decide over the hypothesis.
X[bar]₁-X[bar]₂ ± [tex]Z_{1-\alpha /2}[/tex]*(√(σ₁²/n₁+σ₂²/n₂)
[tex]Z_{1-\alpha /2}[/tex] = [tex]Z_{0,975}[/tex] = 1,96
[28900-30300 ± 1.96*(√((2300)²/15+(2100)²/15)]
[-1400 ± 1576,15]
[-2976.15;176,15]
Since I've approximated to two decimal units in the intermediate calculations, the values differ slightly, but the interval is:
(–2975, 175)Now, since the 0 is contained in the Confidence interval, the decision is to not reject the null hypothesis. In other words, the difference in average salaries between the two districts is cero.
I hope it helps!
The appropriate set of hypotheses (H0, H1) is B. µ1 – µ2 ≠ 0, µ1 – µ2 = 0.
How to illustrate the hypothesis?The null hypothesis in a test predicts that there's no relationship between the variables. From the information, the null and alternative hypothesis are reflected by µ1 – µ2 ≠ 0, µ1 – µ2 = 0.
The confidence interval will be:
= (28900 - 30300) + 2.13 × ✓(2 × 4850000/15)
= = (28900 - 30300) + 2.13 × 804.15
= 325
Also, (28900 - 30300) - 2.13 × ✓(2 × 4850000/15)
= = (28900 - 30300) - 2.13 × 804.15
= -3125
The confidence interval will be (–3125, 325).
Lastly, the salaries of the teachers from the two districts isn't different as it's equal since the confidence interval has negative and positive values.
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An agricultural field trial compares the yield of two varieties of tomatoes for commercial use – one is the common variety used by farmers and the other is a new hybrid. The researchers plant each type of tomato on one half of each of 10 plots. At harvest time the difference in the yield of the hybrid tomato minus the common variety in pounds per plant (ppp) is calculated for each plot. The average difference is 1.05 ppp with a standard deviation of 1.23 ppp. In testing whether the hybrid tomato has a higher yield with a significance level of α = 0.05, what is the p-value? You may assume the conditions to carry out the relevant hypothesis test are satisfied.
Answers
Answer:
p-value 0.0124
Step-by-step explanation:
Hello!
In this experiment, 10 plots were randomly sampled and each plot was divided in a half, one half was planted with the common variety of tomato and the other half was planted with the new hybrid variety. At harvest time it was measured the pounds per plant harvested and the variable difference was established. This study variable is defined as the difference between the pounds per plant of the hybrid variety minus the pound per plant of the common variety. Symbolized: Xd: Xnew - Xcommon.
The purpose of this experiment is to test whether there is a difference between the yield of both species. This is a classic example of a paired test, in which you want to put two dependent samples to test. The easiest way to recognize this type of test, if it is not specified in the problem, is that both variables are measured in the same sampling unit. In this case, the sampling unit is "one plot" that was divided and both verities of tomatoes were planted on it.
This test is also called "paired samples T-test" and as a statistic, you have to use the Student t.
The data given is:
sample: 10 plots
sample mean: Xd[bar]: 1.05ppp
sample standard deviation Sd: 1.23ppp
If the hybrid has a higher yield, that would mean that the difference between them would be positive if the difference is positive, its safe to assume that the population mean of the difference will be positive as well. Symbolically: μd>0
With this, you can state the hypothesis as:
H₀:μd≤0
H₁:μd>0
α: 0.05
t= Xd[bar] - μd ~ [tex]t_{n-1}[/tex]
Sd/n
t[tex]t_{obs}[/tex]= (1.05 - 0) / (1.23/√10) = 2.69
This is a one-tailed test with critical value [tex]t_{9;0.95} = 1.83[/tex]
the p-value for this test is also one tailed.
p([tex]t_{9}[/tex]≥2.69) = 1-P([tex]t_{9}[/tex]<2.69) = 1 - 0.9876 = 0.0124
The decision is to reject the null hypothesis.
I hope it helps!
Let X equal the thickness of spearmint gum manufactured for vending machines. Assume that the distribution of X is N(mu, sigma^2). The target thickness is 7.5 hundreds of an inch. We shall test the null hypothesis H_0: mu=7.5 against a two-sided alternative hypothesis, using 10 observations. Let m denote the sample mean and S^2 denote the sample variance.
Define the test statistic in terms of m and s.
Answers
Answer:
[tex]t=\frac{m-7.5}{\frac{s}{\sqrt{10}}}=\sqrt{10} (\frac{m-7.5}{s})[/tex]
Step-by-step explanation:
1) Notation
n=10 represent the sample size
[tex]\bar X=m[/tex] represent the sample mean
[tex]s[/tex] represent the sample standard deviation
m represent the margin of error
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The null hypothesis attempts "to show that no variation exists between variables or that a single variable is no different than its mean"
The alternative hypothesis "is the hypothesis used in hypothesis testing that is contrary to the null hypothesis"
2) State the null and alternative hypotheses.
We need to conduct a hypothesis in order to determine if the mean for the population is 7.5 or no, the system of hypothesis would be:
Null hypothesis:[tex]\mu =7.5[/tex]
Alternative hypothesis:[tex]\mu \neq 7.5[/tex]
We don't know the population deviation, so for this case we can use the t test to compare the actual mean to the reference value, and the statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
3) Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]t=\frac{m-7.5}{\frac{s}{\sqrt{10}}}=(\sqrt{10})\frac{m-7.5}{s}[/tex]
and we have our statistic in terms of m (mean) and the sample standard deviation s.
A cable network offers members a Basic plan for $7.31 per month. For $3.00 more per month, the cable network offers a Standard plan, which includes HD movies. During one week, 280 new subscribers paid a total of $2286.80 for their plans. How many Basic plans and how many Standard plans were purchased?
Answers
Final answer:
By setting up a system of equations based on the given information, we find that 200 Basic plans and 80 Standard plans were purchased by the new subscribers.
Explanation:
A cable network offers two types of plans: a Basic plan for $7.31 per month and a Standard plan which costs $3.00 more per month, totaling $10.31. To find out how many Basic and Standard plans were purchased by 280 new subscribers, who altogether paid $2286.80, we can set up a system of equations. Let x represent the number of Basic plans and y represent the number of Standard plans.
The total amount of plans purchased: x + y = 280
The total amount paid: 7.31x + 10.31y = 2286.80
Solving this system of equations:
Step 1: Multiply the first equation by 7.31 to make the x coefficients equal: 7.31x + 7.31y = 2046.80
Step 2: Subtract this equation from the second equation to eliminate x and solve for y: 3y = 240, thus y = 80. Therefore, 80 Standard plans were purchased.
Step 3: Substitute y back into one of the original equations to solve for x: x + 80 = 280, thus x = 200. Therefore, 200 Basic plans were purchased.
So, there were 200 Basic plans and 80 Standard plans purchased by the new subscribers.
Which of the following equations could be the result of using the comparison method to solve the system shown? x + y = 5 2x + y = 7
Answers
Answer:
so (x,y) = (2,3)
Step-by-step explanation:
we have
x +y= 5 ------ equation 1
2x+y= 7-------equation 2
Multiplying equation--1 by 2, we will have equation---3
2x +2y= 10-----equation 3
now subtracting equation 2 from equation 3
2x + 2y= 10
2x + y = 7
y = 3
adding value of y in equation 1
x + y = 5
x= 5 -y
x= 5 - 3
x =2
so (x,y) = (2,3)
Answer: the correct answer is
5 - x = 7 - 2x
Step-by-step explanation:
Wire electrical-discharge machining (WEDM) is a process used to manufacture conductive hard metal components. It uses a continuously moving wire that serves as an electrode. Coating on the wire electrode allows for cooling of the wire electrode core and provides an improved cutting performance. An article gave the following sample observations on total coating layer thickness (in µm) of eight wire electrodes used for WEDM. 21 16 29 36 42 25 24 25
Calculate a 99% CI for the standard deviation of the coating layer thickness distribution. (Round your answers to two decimal places.) , Is this interval valid whatever the nature of the distribution? Explain.
Yes, there are enough data points for this interval to be valid.
No, validity of this interval requires that coating layer thickness be, at least approximately, normally distributed.
Answers
To calculate a 99% confidence interval for the standard deviation of the coating layer thickness distribution, use the sample variance, sample size, and chi-square distribution. The coating layer thickness should be approximately normally distributed for the confidence interval to be valid.
Explanation:To calculate a 99% confidence interval for the standard deviation of the coating layer thickness distribution, we can use the chi-square distribution. First, we need to calculate the sample variance and sample size. Then, we can use the chi-square distribution table to find the critical values for a 99% confidence interval with (n-1) degrees of freedom. Finally, we can calculate the confidence interval using the formula:
CI = sqrt((n - 1) * s^2 / X^2)
where CI is the confidence interval, n is the sample size, s^2 is the sample variance, and X^2 is the critical value from the chi-square distribution.
In this case, the coating layer thickness should be approximately normally distributed for the confidence interval to be valid.
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In the given case, the answer No, validity of this interval requires that coating layer thickness be, at least approximately, normally distributed.
To calculate the 99% confidence interval (CI) for the standard deviation of the coating layer thickness distribution, we can use the chi-square distribution.
Given that the sample size is small (n = 8), we'll use the chi-square distribution with n - 1 = 7 degrees of freedom.
The formula for the confidence interval is:
[tex]\[ \left( \sqrt{\frac{(n-1)S^2}{\chi^2_{\alpha/2, n-1}}}, \sqrt{\frac{(n-1)S^2}{\chi^2_{1-\alpha/2, n-1}}} \right) \][/tex]
Given the sample observations: 21, 16, 29, 36, 42, 25, 24, 25, we first need to calculate the sample standard deviation:
[tex]\[ S = \sqrt{\frac{\sum_{i=1}^{n}(X_i - \bar{X})^2}{n-1}} \][/tex]
Let's perform the calculations:
[tex]\[ S = \sqrt{\frac{(21-27.625)^2 + (16-27.625)^2 + \ldots + (25-27.625)^2}{7}} \][/tex]
[tex]\[ S \approx 9.38 \][/tex]
Now, plug the values into the formula:
[tex]\[ \left( \sqrt{\frac{7 \times 9.38^2}{18.48}}, \sqrt{\frac{7 \times 9.38^2}{2.17}} \right) \][/tex]
[tex]\[ \left( \sqrt{26.97}, \sqrt{229.78} \right) \][/tex]
[tex]\[ \left( 5.19, 15.15 \right) \][/tex]
Therefore, the 99% confidence interval for the standard deviation of the coating layer thickness distribution is approximately [tex]\( (5.19, 15.15) \)[/tex] micrometers.
Therefore, the answer is: No, validity of this interval requires that coating layer thickness be, at least approximately, normally distributed.
In a recent poll, 600 people were asked if they liked soccer, and 72% said they did. Based on this, construct a 99% confidence interval for the true population proportion of people who like soccer.
As in the reading, in your calculations:
--Use z = 1.645 for a 90% confidence interval
--Use z = 2 for a 95% confidence interval
--Use z = 2.576 for a 99% confidence interval.
Give your answers as decimals, to 4 decimal places.
Out of 500 people sampled, 80 had kids. Based on this, construct a 90% confidence interval for the true population proportion of people with kids.
As in the reading, in your calculations:
--Use z = 1.645 for a 90% confidence interval
--Use z = 2 for a 95% confidence interval
--Use z = 2.576 for a 99% confidence interval.
Give your answers to three decimals
Give your answers as decimals, to three decimal places.
Answers
Answer:
(0.6728, 0.7672)
(0.1330, 0.1870)
Step-by-step explanation:
Given that in a recent poll, 600 people were asked if they liked soccer, and 72% said they did.
Std error = [tex]\sqrt{\frac{pq}{n} } \\=\sqrt{\frac{0.72(0.28)}{600} } \\=0.0183[/tex]
Margin of error for 99% we would use the value [tex]z = 2.576[/tex]
Margin of error = [tex]2.576*SE\\=2.576*0.0183\\=0.0472[/tex]
Confidence interval lower bound = [tex]0.72-0.0472=0.6728[/tex]
Upper bound = [tex]0.72+0.0472=0.7672[/tex]
99% confidence interval for the true population proportion of people who like soccer.=(0.6728, 0.7672)
b) n =500
Sample proportion p=[tex]\frac{80}{500} =0.16[/tex]
Margin of error for 90% = [tex]1.645*\sqrt{\frac{0.16*0.84}{500} } \\\\=0.0270[/tex]
90% confidence interval for the true population proportion of people with kids. =[tex](0.16-0.0270, 0.16+0.0270)\\=(0.1330, 0.1870)[/tex]
A magazine currently has 8700 subscribers for its online web version. It is adding members at the rate of R(t) = 190e0.03t subscribers per month. If the proportion of members who remain subscribed t months from now is S(t) = e−0.06t, how many online subscribers will the magazine have three years from now? (Round your answer to the nearest integer.)
Answers
Number of subscriber the magazine will have after 3 years from now approximately be 8767
Solution:
Given that magazine currently has 8700 subscribers for its online web version
[tex]\begin{array}{l}{\mathrm{R}(\mathrm{t})=190 \mathrm{e}^{0.03 \mathrm{t}} \text { subscribers/month }} \\\\ {\mathrm{S}(\mathrm{t})=\mathrm{e}^{-0.06 \mathrm{t}}}\end{array}[/tex]
After 3 years, time(t) = 36 month
Total number of subscribers after 3 years from now :
Substitute "t" = 36
[tex]\begin{array}{l}{\mathrm{R}(36)=190 \mathrm{e}^{0.03 \times(36)}=190 \times(2.944)} \\\\ {\mathrm{R}(36) \approx 560} \\\\ {\mathrm{S}(36)=\mathrm{e}^{-0.06 \times(36)}=0.12}\end{array}[/tex]
Subscribers remaining = 0.12 x 560 = 67.2
The magazine currently has 8700 subscribers
Added Subscriber = 8700 + 560 = 9260
Remaining Subscriber = 8700 + 67.2 = 8767.2
Therefore number of subscriber the magazine will have after 3 years from now approximately be 8767
A statistical program is recommended.
A random sample of soil specimens was obtained, and the amount of organic matter (%) in the soil was determined for each specimen, resulting in the accompanying data.
1.12 5.09 0.97 1.59 4.60 0.32 0.55 1.45
0.18 4.47 1.20 3.50 5.02 4.67 5.22 2.69
3.96 3.17 3.03 2.21 0.69 4.47 3.31 1.17
0.79 1.17 1.57 2.62 1.66 2.05
The values of the sample mean, sample standard deviation, and (estimated) standard error of the mean are 2.484, 1.611, and 0.294, respectively. Does this data suggest that the true average percentage of organic matter in such soil is something other than 3%? Carry out a test of the appropriate hypotheses at significance level 0.10. [Note: A normal probability plot of the data shows an acceptable pattern in light of the reasonably large sample size.]
Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to three decimal places.)
t=
P-value=
Answers
Answer:
Which is the output of the formula =AND(12>6;6>3;3>9)?
A.
TRUE
B.
FALSE
C.
12
D.
9
Step-by-step explanation:
A flagpole that is 40ft in height casts a shadow that is 25ft in length. If a nearby building is 200ft tall, and the sun hits it at the same angle as the tree, what is the length of the building’s shadow?
Answers
Answer:
I would assume that the shadow length is proportional to the height of the flagpole. Hence as the building is 5 times higher than the flagpole, the length of the shadow would be 5 times higher, 25x5= 125 ft.
Step-by-step explanation:
The solution is 125 feet
The length of the building's shadow is 125 feet
What is Proportion?
The proportion formula is used to depict if two ratios or fractions are equal. The proportion formula can be given as a: b::c : d = a/b = c/d where a and d are the extreme terms and b and c are the mean terms.
Given data ,
Let the length of the building's shadow be = A
Now , the equation will be
Let the height of the flagpole be = 40 feet
Let the length of the shadow of flagpole be = 25 feet
Let the height of the building be = 200 feet
So , the proportion between the height and length is given by
Height of the flagpole / length of the shadow of flagpole = height of the building / length of the building's shadow
Substituting the values in the equation , we get
40 / 25 = 200 / A
On simplifying the equation , we get
1.6 = 200 / A
Multiply by A on both sides of the equation , we get
1.6A = 200
Divide by 1.6 on both sides of the equation , we get
A = 200/1.6
The value of A = 125 feet
Therefore , the value of A is 125 feet
Hence , the length of the shadow is 125 feet
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In a random sample of 150 customers of a high-speed Internetprovider, 63 said that their service had been interrupted one ormore times in the past month.A) Find a 95% confidence interval for the proportion of customerswhose service was interrupted one or more times in the pastmonth.B) Find a 99% confidence interval for the proportion of customerswhose service was interrupted one or more times in the pastmonth.C) Find the sample size needed for a 95% confidence interval tospecify the proportion to within -+ 0.05.D) Find the sample size needed for a 99% confidence interval tospecify the proportion to within -+ 0.05.
Answers
Answer:
a) The 95% confidence interval would be given by (0.341;0.499)
b) The 99% confidence interval would be given by (0.316;0.524)
c) n=335
d)n=649
Step-by-step explanation:
1) Notation and definitions
[tex]X_{IS}=63[/tex] number of high speed internet users that had been interrupted one or more times in the past month.
[tex]n=150[/tex] random sample taken
[tex]\hat p_{IS}=\frac{63}{150}=0.42[/tex] estimated proportion of high speed internet users that had been interrupted one or more times in the past month.
[tex]p_{IS}[/tex] true population proportion of high speed internet users that had been interrupted one or more times in the past month.
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The population proportion have the following distribution
[tex]p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})[/tex]
1) Part a
In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]. And the critical value would be given by:
[tex]t_{\alpha/2}=-1.96, t_{1-\alpha/2}=1.96[/tex]
The confidence interval for the mean is given by the following formula:
[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]
If we replace the values obtained we got:
[tex]0.42 - 1.96\sqrt{\frac{0.42(1-0.42)}{150}}=0.341[/tex]
[tex]0.42 + 1.96\sqrt{\frac{0.42(1-0.42)}{150}}=0.499[/tex]
The 95% confidence interval would be given by (0.341;0.499)
2) Part b
In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by [tex]\alpha=1-0.99=0.01[/tex] and [tex]\alpha/2 =0.005[/tex]. And the critical value would be given by:
[tex]t_{\alpha/2}=-2.58, t_{1-\alpha/2}=2.58[/tex]
The confidence interval for the mean is given by the following formula:
[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]
If we replace the values obtained we got:
[tex]0.42 - 2.58\sqrt{\frac{0.42(1-0.42)}{150}}=0.316[/tex]
[tex]0.42 + 2.58\sqrt{\frac{0.42(1-0.42)}{150}}=0.524[/tex]
The 99% confidence interval would be given by (0.316;0.524)
3) Part c
The margin of error for the proportion interval is given by this formula:
[tex] ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex] (a)
And on this case we have that [tex]ME =\pm 0.05[/tex] and we are interested in order to find the value of n, if we solve n from equation (a) we got:
[tex]n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}[/tex] (b)
And replacing into equation (b) the values from part a we got:
[tex]n=\frac{0.42(1-0.42)}{(\frac{0.05}{1.96})^2}=374.32[/tex]
And rounded up we have that n=335
4) Part d
The margin of error for the proportion interval is given by this formula:
[tex] ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex] (a)
And on this case we have that [tex]ME =\pm 0.05[/tex] and we are interested in order to find the value of n, if we solve n from equation (a) we got:
[tex]n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}[/tex] (b)
And replacing into equation (b) the values from part a we got:
[tex]n=\frac{0.42(1-0.42)}{(\frac{0.05}{2.58})^2}=648.599[/tex]
And rounded up we have that n=649
Please help if so thank you And explain
Answers
Answer:
x = 34°
Step-by-step explanation:
Given AC and BD are perpendicular bisectors, we can say that at point E, there are 4 right angles [perpendicular bisectors intersect to create 4 90 degree angles].
Now, if we look at the triangle AED, we know that it is a right triangle, meaning that angle E is a right angle.
Also,
We know sum of 3 angles in a triangle is 180 degrees. Thus, we can write:
∠A + ∠E + ∠D = 180
Note: Angle A and Angle D are just the half part of the diagram. More exactly we can write:
∠EAD + ∠ADE + ∠DEA = 180
Given,
∠EAD = 56
∠DEA = 90
We now solve:
∠EAD + ∠ADE + ∠DEA = 180
56 + ∠ADE + 90 = 180
146 + ∠ADE = 180
146 + x = 180
x = 180 - 146
x = 34°
In the exercise, X is a binomial variable with n = 6 and p = 0.2. Compute the given probability. Check your answer using technology. HINT [See Example 2.] (Round your answer to five decimal places.)P(3 ≤ X ≤ 5)
Answers
Answer:
[tex]P(3\leq X \leq 5)=0.09882[/tex]
Step-by-step explanation:
1) Previous concepts
The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".
2) Solution to the problem
Let X the random variable of interest, on this case we now that:
[tex]X \sim Binom(n=6, p=0.2)[/tex]
The probability mass function for the Binomial distribution is given as:
[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]
Where (nCx) means combinatory and it's given by this formula:
[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]
And we want to find this probability:
[tex]P(3 \leq x \leq 5)=P(X=3)+P(X=4)+P(X=5)[/tex]
[tex]P(X=3)=(6C3)(0.2)^3 (1-0.2)^{6-3}=0.08192[/tex]
[tex]P(X=4)=(6C4)(0.2)^4 (1-0.2)^{6-4}=0.01536[/tex]
[tex]P(X=5)=(6C5)(0.2)^5 (1-0.2)^{6-5}=0.001536[/tex]
[tex]P(3 \leq x \leq 5)=0.08192+0.01536+0.001536=0.09882[/tex]
The drug OxyContin (oxycodone) is used to treat pain, but it is dangerous because it is addictive and can be lethal. In clinical trials, 227 subjects were treated with OxyContin and 52 of them developed nausea (based on data from Purdue Pharma L.P.). Use a 0.05 significance level to test the claim that more than 20% of OxyContin users develop nausea. Does the rate of nausea appear to be too high?
Answers
Answer:
Step-by-step explanation:
According to the record of the registrar's office at a state university, 35% of the students arefreshman, 25% are sophomore, 16% are junior, and the rest are senior. Among the freshmen,sophomores, juniors, and seniors, the portion of students who live in the dormitory are,respectively, 80%, 60%, 30%, and 20%. What is the probability that a randomly selected studentis a freshman who lives in a dormitory?
A) 0.28
B) 0.32
C) 0.52
D) 0.38
Answers
Answer:
the probability is 0.28
Step-by-step explanation:
using Bayes's theorem
P(A|B)=P(A∩B)/P(B)
where
P(A∩B) = probability that events A and B happen
P(A|B) = probability that event A happen if B already happened
P(B)= probability of event B
therefore
P(A∩B)=P(A|B)*P(B)
if event A= selection of a student that lives in a dormitory and event B = selection of a freshmen student
P(A|B) = 0.8 (live in a dormitory knowing that is a freshmen student )
P(B) = 0.35 (freshmen student)
P(A∩B)=P(A|B)*P(B) = 0.8* 0.35 =0.28
The probability that a randomly selected student is a freshman who lives in a dormitory is 0.28. So the correct option is A.
Explanation:To find the probability that a randomly selected student is a freshman who lives in a dormitory, we need to multiply the probability that a student is a freshman by the probability that a freshman lives in the dormitory. According to the given information, 35% of the students are freshmen, and among them, 80% live in the dormitory.
We calculate this probability using the formula:
Probability of (Freshman who lives in dorm) = Probability of (Freshman) × Probability of (Dorm | Freshman)
Probability of (Freshman who lives in dorm) = 0.35 × 0.80
Probability of (Freshman who lives in a dorm) = 0.28
Thus, the correct answer is A) 0.28.
An insurance company sets up a statistical test with a null hypothesis that the average time for processing a claim is 7 days, and an alternative hypothesis that the average time for processing a claim is greater than 7 days.
After completing the statistical test, it is concluded that the average time exceeds 7 days.
However, it is eventually learned that the mean process time is really 9 days.
What type of error occurred in the statistical test?
Answers
Answer:
There was no error
Step-by-step explanation:
There are two types of statistical errors, the type 1 error and the type 2 error. In this case we refute the null hypothesis when the hypothesis is, in fact, false, because the mean process is 9 days instead of 7. Therefore we made no errors.
If the null hypothesis were True, a type 1 error would have ocurred. If the null hypothesis were false and we didnt refute it, then a type 2 error would have ocurred.
A sample of 20 cigarettes is tested to determine nicotine content and the average value observed was 1.2 mg. Compute a 99 percent two-sided confidence interval for the mean nicotine content of a cigarette if it is known that the standard deviation of a cigarette’s nicotine content is σ = .2 mg.
Answers
Answer: [tex]1.0848<\mu<1.3152[/tex]
Step-by-step explanation:
Confidence interval for population mean is given by :-
[tex]\overline{x}-z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}< mu< \overline{x}+ z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}[/tex]
, where [tex]z_{\alpha/2}[/tex] = two -tailed z-value for [tex]{\alpha[/tex] (significance level)
n= sample size .
[tex]\sigma[/tex] = Population standard deviation.
[tex]\overline{x}[/tex] = Sample mean
By considering the given information , we have
[tex]\sigma=0.2\text{ mg}[/tex]
[tex]\overline{x}=1.2\text{ mg}[/tex]
n= 20
[tex]\alpha=1-0.99=0.01[/tex]
Using z-value table ,
Two-tailed Critical z-value : [tex]z_{\alpha/2}=z_{0.005}=2.576[/tex]
The 99 percent two-sided confidence interval for the mean nicotine content of a cigarette will be :-
[tex]1.2- (2.576)\dfrac{0.2}{\sqrt{20}}<\mu<1.2+ (2.576)\dfrac{0.2}{\sqrt{20}}\\\\=1.2- 0.1152<\mu<1.2+ 0.1152\\\\=1.0848<\mu<1.3152 [/tex]
Hence, the 99 percent two-sided confidence interval for the mean nicotine content of a cigarette: [tex]1.0848<\mu<1.3152[/tex]
In the triangle above, which angle is smallest in measure?
A. ∠A
B. ∠B
C. ∠C
D. All three angles are congruent.
Answers
Answer:
A. ∠A
Step-by-step explanation:
The shortest side is BC. The smallest angle is opposite the shortest side, so is angle A.
∠A is the smallest in measure.
To test μ for an x distribution that is mound-shaped using sample size n ≥ 30, how do you decide whether to use the normal or Student's t distribution?
A.If σ is known, use the standard normal distribution. If σ is unknown, use the Student's t distribution with n – 1 degrees of freedom.
B. If σ is unknown, use the standard normal distribution. If σ is known, use the Student's t distribution with n – 1 degrees of freedom.
C. If σ is known, use the standard normal distribution. If σ is unknown, use the Student's t distribution with n degrees of freedom.
D. For large samples we always the standard normal distribution.
Answers
Answer:
The correct option is A) If σ is known, use the standard normal distribution. If σ is unknown, use the Student's t distribution with n – 1 degrees of freedom.
Step-by-step explanation:
Consider the provided information.
The t-distribution of the Student is a distribution of probability that is used when when the sample size is small and/or when the population variance is unknown to estimate population parameters.
The number of independent observations is equal to the sample size minus one when calculating a mean score or a proportion from a single sample.
Since µ and σ determine the shape of the distribution so we use standard normal distribution if σ is known.
Hence, the correct option is A) If σ is known, use the standard normal distribution. If σ is unknown, use the Student's t distribution with n – 1 degrees of freedom.
Ms. Taylor's students give cards to each of
the 3 class parent helpers. There are
24 cards. How many cards will each helper
get if the students give an equal number of
cards to each helper?
Answers
To determine the number of cards each class parent helper will receive, divide the total number of cards (24) by the number of helpers (3), resulting in each helper receiving 8 cards.
Ms. Taylor's students are giving out cards to each of the 3 class parent helpers equally, and there are a total of 24 cards. To find out how many cards each helper gets, we can divide the total number of cards by the number of helpers.
Step 1: Determine the total number of cards: 24.
Step 2: Determine the number of class parent helpers: 3.
Step 3: Divide the total number of cards by the number of helpers: 24 ÷ 3.
Step 4: Calculate: 24 ÷ 3 = 8.
So, each class parent helper will receive 8 cards.
Using lengths a, b, and c in the right triangle shown, how are the trigonometric functions of θ defined?
Answers
Answer:
Step-by-step explanation:
From the diagram
b is the length of side adjacent to the angle (x) in the question
a is the length of side opposite to the angle
c is the length of hypotenuse
θ represents the measure of the angle in either degree or radians
From the diagram
Cosecant (csc θ) = c/a
Secant (sec θ) = c/b
Cotangent (cot θ) = b/a
The trigonometric functions in a right triangle are defined as the ratios of the sides: sine is opposite over hypotenuse, cosine is adjacent over hypotenuse, and tangent is opposite over adjacent.
Explanation:In a right triangle with sides of lengths a, b, and c (where c is the hypotenuse) and an angle θ, the trigonometric functions are defined as follows:
Sine (sin θ) is defined as the length of the side opposite θ (a) divided by the length of the hypotenuse (c), or sin θ = a/c.Cosine (cos θ) is defined as the length of the side adjacent θ (b) divided by the hypotenuse (c), or cos θ = b/c.Tangent (tan θ) is defined as the length of the side opposite θ (a) divided by the adjacent side (b), or tan θ = a/b.These functions can easily be remembered using the mnemonic "SOHCAHTOA" - Sine equals Opposite over Hypotenuse, Cosine equals Adjacent over Hypotenuse, and Tangent equals Opposite over Adjacent.
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At 95% confidence, how large a sample should be taken to obtain a margin of error of .03 for the estimation of the population proportion? Assume past data are not available for developing a planning value for p*.
Answers
Answer:
The sample should be 1,068.
Step-by-step explanation:
Consider the provided information.
Confidence level is 95% and margin of error is 0.03.
Thus,
1-α=0.95
α=0.05, E=0.03 and planning value [tex]\hat p=0.5[/tex]
Formula to calculate sample size is: [tex]n=\frac{\hat p(1-\hat p)(z_{\alpha/2})^2}{E^2}[/tex]
From the table we can find:
[tex]z_{\alpha/2}=z_{0.05/2}\\z_{0.025}=1.96[/tex]
Substitute the respective values in the above formula we get:
[tex]n=\frac{0.5(0.5)(1.96)^2}{(0.03)^2}[/tex]
[tex]n=\frac{0.25(1.96)^2}{(0.03)^2}\approx 1067.111[/tex]
Hence, the sample should be 1,068.
Craig measured these three rectangles.
Answers
Answer:
Step-by-step explanation:
Perimeter of a rectangle is expressed as (2 length + 2 width) = 2(L + W)
A) The length of rectangle A is y + 1
The width of rectangle A is x
Perimeter of rectangle A = 2(y + 1 + x) = 2y + 2 + 2x
= 2x + 2y + 2
The length of rectangle B is 2x - 2y
The width of rectangle B is x + 1
Perimeter of rectangle B = 2(2x - 2y+ x + 1) = 4x - 4y + 2x + 2) = 4x + 2x - 4y + 2
= 6x - 4y + 2
The length of rectangle C is 3x + 3y
The width of rectangle C is 2x - 3
Perimeter of rectangle C = 2(3x + 3y + 2x - 3) = 6x + 6y + 4x - 6) =
(6x + 6y + 4x - 6)
= 10x + 6y - 6
B) The combined perimeters will be the sum if perimeter of rectangle A, perimeter of rectangle B and perimeter of rectangle C. It becomes
2x + 2y + 2 + 6x - 4y + 2 + 10x + 6y - 6
Collecting like terms
2x + 6x + 10x + 2y + 6y - 4y + 2 + 2 - 6
The combined perimeter = 18x + 4y - 2
Answer:
In bold below.
Step-by-step explanation:
A.
2x + 2(y + 1)
= 2x + 2y + 2.
2(2x - 2y) + 2(x + 1)
= 4x - 4y + 2x + 2
= 6x - 4y + 2.
2(3x + 3y ) + 2(2x - 3)
= 6x + 6y + 4x - 6
= 10x + 6y - 6.
B.
2x + 2y + 2 + 10x + 6y - 6 + 6x - 4y + 2
= 18x + 4y - 2.
The area of a rectangular plot 32 feet long and 25 feet wide will be doubled by adding an equal width to each side of the plot. Which equation can be used to find this added width?
(x + 32)(x + 25) = 1600
(2x + 32)(2x + 25) = 1600
(2x + 32)(x + 25) = 800
(x + 32)(x + 25) = 800
Answers
Final answer:
To find the added width of the rectangular plot, set up the equation (32 + x)(25 + x) = 2(32)(25) based on the fact that the area is doubled. Solving this equation will give the added width.
Explanation:
To find the added width of the rectangular plot, we need to determine the increase in length and width that will double the area. The original dimensions of the plot are 32 feet long and 25 feet wide. Since equal width is added to each side, the new length and width will be 32 + x and 25 + x, respectively. To find x, we can set up the equation (32 + x)(25 + x) = 2(32)(25) using the fact that the area is doubled. Solving this equation will give us the added width.
Expanding (32 + x)(25 + x), we get 800 + 57x + x^2 = 1600. Rearranging and simplifying, we get x^2 + 57x - 800 = 0. We can then solve this quadratic equation using factoring or the quadratic formula to find the value of x.
The correct equation to use in order to find the added width is (x + 32)(x + 25) = 800.
The Greater Pittsburgh Area Chamber of Commerce wants to estimate the mean time workers who are employed in the downtown area spend getting to work. A sample of the 15 workers reveals the following number of minutes spent traveling. 29 38 38 33 38 21 45 34 40 37 37 42 30 29 35Develop a 98 percent confidence interval for the population mean. Interpret the results.
Answers
Answer:
98% Confidence interval: (31.74,38.4)
Step-by-step explanation:
We are given the following data set:
29, 38, 38, 33, 38, 21, 45, 34, 40, 37, 37, 42, 30, 29, 35
Formula:
[tex]\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}[/tex]
where [tex]x_i[/tex] are data points, [tex]\bar{x}[/tex] is the mean and n is the number of observations.
[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]
[tex]Mean =\displaystyle\frac{526}{15} = 35.07[/tex]
Sum of squares of differences = 506.93
[tex]S.D = \sqrt{\displaystyle\frac{506.93}{14}} = 6.02[/tex]
98% Confidence interval:
[tex]\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}[/tex]
Putting the values, we get,
[tex]t_{critical}\text{ at degree of freedom 14 and}~\alpha_{0.02} = \pm 2.145[/tex]
[tex]35.07 \pm 2.145(\frac{6.02}{\sqrt{15}} ) = 35.07 \pm 3.33 = (31.74,38.4)[/tex]
Thus, there is 98% confidence that the population mean number of minutes spent traveling by workers is between 31.74 mins and 38.40
A study of 20 worldwide financial institutions showed the correlation between their assets and pretax profit to be 0.81.State the decision rule for 0.025 significance level: H0: rho ≤ 0; H1: rho > 0. (Round your answer to 3 decimal places.)Compute the value of the test statistic. (Round your answer to 2 decimal places.)Can we conclude that the correlation in the population is greater than zero? Use the 0.025 significance level.
Answers
Answer:
t=5.86
We can conclude that the population correlation between their assets and pretax profit is higher than 0 at the significance level provided.
Step-by-step explanation:
n= 20 random sample taken
r=0.81 correlation coeffcient obtained
[tex]\alpha=0.025[/tex] significance level obtained
1) System of hypothesis
The system of hypothesis given are:
Null hypothesis :[tex]\rho \leq 0[/tex]
Alternative hypothesis: [tex] \rho >0[/tex]
2) Calculate the statistic
The statistic in order to test an hypothesis for the correlation coefficient is given by:
[tex]t =\frac{r\sqrt{n-2}}{\sqrt{1-r^2}}[/tex]
This statistic follows a t distribution with n-2 degrees of freedom
If we replace the values given we got:
[tex]t =\frac{0.81\sqrt{20-2}}{\sqrt{1-(0.81^2)}}=5.86[/tex]
3) P value
For this case w eneed to calculate first the degrees of freedom
[tex]df=n-2=20-2=18[/tex]
And then analyzing the alternative hypothesis we can calculate the p value on this way:
[tex]p_v =P(t_{18} >5.86) =1-P(t_{18} <5.86)=1-0.99999=7.51x10^{-6}[/tex]
Since the P-value is smaller than the significance level, we have enough evidence to reject the null hypothesis in favor of the alternative. We conclude "there is sufficient evidence at the significance level to conclude that there is a linear relationship in the population between the two variables analyzed."
The decision rule for a 0.025 significance level intends to reject the null hypothesis if the p-value from the test statistic is less than 0.025. Using a standard Z test to compute the test statistic, we substitute the given values to find the result. As the p-value is 0.026, which is greater than the 0.025 significance level, we do not reject the null hypothesis, indicating insufficient evidence to conclude that the population correlation is greater than zero.
Explanation:To answer your question, let's first state the decision rule for a 0.025 significance level. Considering your null hypothesis (H0: rho ≤ 0) and the alternative hypothesis (H1: rho > 0), we want to reject the null hypothesis if the p-value from the test statistic is less than 0.025.
Since we aren't given a specific formula or context about how the test statistic should be computed, I'll assume that it's via a standard Z test using the formula:
[tex]Z = (r\times \sqrt(n-2))/(\sqrt(1-r^2)),[/tex]where r is the sample correlation (0.81 in this case), and n is the number of data points (20 in this case). Let's substitute these values into the formula to find the test statistic.
Based upon the p-value from your computing software as 0.026 and the significance level being 0.025, we do not reject the null hypothesis because the p-value is greater than the significance level. In other words, there is insufficient evidence at the 0.025 significance level to conclude that the population correlation is greater than zero.
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Let the number of chocolate chips in a certain type of cookie have a Poisson distribution. We want the probability that a cookie of this type contains at least two chocolate chips to be greater than 0.99. Find the smallest value of the mean that the distribution can take.
Answers
Answer:
[tex]\lambda \geq 6.63835[/tex]
Step-by-step explanation:
The Poisson Distribution is "a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known constant mean rate and independently of the time since the last event".
Let X the random variable that represent the number of chocolate chips in a certain type of cookie. We know that [tex]X \sim Poisson(\lambda)[/tex]
The probability mass function for the random variable is given by:
[tex]f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...[/tex]
And f(x)=0 for other case.
For this distribution the expected value is the same parameter [tex]\lambda[/tex]
[tex]E(X)=\mu =\lambda[/tex]
On this case we are interested on the probability of having at least two chocolate chips, and using the complement rule we have this:
[tex]P(X\geq 2)=1-P(X<2)=1-P(X\leq 1)=1-[P(X=0)+P(X=1)][/tex]
Using the pmf we can find the individual probabilities like this:
[tex]P(X=0)=\frac{e^{-\lambda} \lambda^0}{0!}=e^{-\lambda}[/tex]
[tex]P(X=1)=\frac{e^{-\lambda} \lambda^1}{1!}=\lambda e^{-\lambda}[/tex]
And replacing we have this:
[tex]P(X\geq 2)=1-[P(X=0)+P(X=1)]=1-[e^{-\lambda} +\lambda e^{-\lambda}[][/tex]
[tex]P(X\geq 2)=1-e^{-\lambda}(1+\lambda)[/tex]
And we want this probability that at least of 99%, so we can set upt the following inequality:
[tex]P(X\geq 2)=1-e^{-\lambda}(1+\lambda)\geq 0.99[/tex]
And now we can solve for [tex]\lambda[/tex]
[tex]0.01 \geq e^{-\lambda}(1+\lambda)[/tex]
Applying natural log on both sides we have:
[tex]ln(0.01) \geq ln(e^{-\lambda}+ln(1+\lambda)[/tex]
[tex]ln(0.01) \geq -\lambda+ln(1+\lambda)[/tex]
[tex]\lambda-ln(1+\lambda)+ln(0.01) \geq 0[/tex]
Thats a no linear equation but if we use a numerical method like the Newthon raphson Method or the Jacobi method we find a good point of estimate for the solution.
Using the Newthon Raphson method, we apply this formula:
[tex]x_{n+1}=x_n -\frac{f(x_n)}{f'(x_n)}[/tex]
Where :
[tex]f(x_n)=\lambda -ln(1+\lambda)+ln(0.01)[/tex]
[tex]f'(x_n)=1-\frac{1}{1+\lambda}[/tex]
Iterating as shown on the figure attached we find a final solution given by:
[tex]\lambda \geq 6.63835[/tex]
The problem pertains to Poisson Distribution in probability theory, focusing on finding the smallest mean (λ) such that the probability of having at least two chocolate chips in a cookie is more than 0.99. This involves solving an inequality using the formula for Poisson Distribution.
Explanation:This problem pertains to the Poisson Distribution, often used in probability theory. In particular, we're looking at the number of events (in this case, the number of chocolate chips) that occur within a fixed interval. Here, the interval under study is a single cookie. The question requires us to find the smallest value of λ (the mean value of the distribution) such that the probability of getting at least two chocolate chips in a cookie is more than 0.99.
Using the formula for Poisson Distribution, the probability of finding k copies of an event is given by:
P(X=k) = λ^k * exp(-λ) / k!
The condition here is that the probability of finding at least 2 copies is more than 0.99. Therefore, you formally need to solve the inequality:
P(X>=2) = 1 - P(X=0) - P(X=1) > 0.99
Substituting the values of P(X=0) and P(X=1) from our standard formula, you will need to calculate and find the smallest value of λ that satisfies this inequality.
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Suppose a random sample of n = 16 observations is selected from a population that is normally distributed with mean equal to 101 and standard deviation equal to 12.
(a) Give the mean and the standard deviation of the sampling distribution of the sample mean x(with a line above). (Enter your standard deviation to two decimal places.)
mean =
standard deviation =
(b) Find the probability that x(line above it) exceeds 107. (Round your answer to four decimal places.)
Answer:
(c) Find the probability that the sample mean deviates from the population mean μ = 101 by no more than 4. (Round your answer to four decimal places.)
Answers
Answer:
101, 3, 0.025, 0.7416
Step-by-step explanation:
Given that a random sample of n = 16 observations is selected from a population that is normally distributed with mean equal to 101 and standard deviation equal to 12.
As per central limit theorem we have
a) Mean of sample mean = [tex]E(\bar x) =\\ \mu =101[/tex]
Std deviation of sample mean = [tex]\frac{\sigma}{\sqrt{n} } =3[/tex]
Mean = 101
Std dev =3.00
b) [tex]P(\bar x >107) = P(Z>\frac{107-101}{3} )\\= P(Z>2) = 0.025[/tex]
c) the probability that the sample mean deviates from the population mean μ = 101 by no more than 4.
=[tex]P(|\bar x-101|) \leq 4\\= P(|z|\leq 1.13)\\= 2(0.3708)\\=0.7416[/tex]
You roll a fair die three times. What is the probability of each of the following?
a) You roll all 4's.
b) You roll all even numbers.
c) None of your rolls gets a number divisible by 2.
d) You roll at least one 2.
e) The numbers you roll are not all 2's.
Answers
This solution calculates the probability of different outcomes when rolling a fair die three times. The results are obtained by defining the successful outcomes versus the total possible outcomes for each specific event.
Explanation:This question is about probability. A fair die has 6 equally likely outcomes. Let's address each part:
You roll all 4's: There's 1 chance in 6 to roll a 4. Since you're rolling the die three times, the probability is (1/6) * (1/6) * (1/6) = 1/216. You roll all even numbers: There are 3 even numbers on a die (2, 4, 6), so the probability is (3/6) * (3/6) * (3/6) = 1/8. None of your rolls gets a number divisible by 2: This is the same as rolling all odd numbers. There are 3 odd numbers on a die (1, 3, 5), so the probability is (3/6) * (3/6) * (3/6) = 1/8. You roll at least one 2: The opposite of this are outcomes without any 2, i.e., combinations of 1, 3, 4, 5, and 6. Thus, subtract combinations without any 2 from total possible combinations: 1 - [(5/6) * (5/6) * (5/6)] = 91/216. The numbers you roll are not all 2's: The only case when this does not happen is when you roll a 2 three times. So, if we subtract the probability of rolling three 2's from 1, we get 1 - (1/6) * (1/6) * (1/6) = 215/216.Learn more about probability here:
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The probability of rolling all 4's is 1/64. The probability of rolling all even numbers is 7/64. The probability of not rolling all 2's is 63/64.
Explanation:a) The probability of rolling all 4's is ⅛3, or 1/64. There is only one way to roll a 4 on a fair die and a total of 6 possible outcomes on each roll, so the probability is 1/6. Since the rolls are independent, the probability of getting a 4 on all three rolls is (1/6)(1/6)(1/6) = 1/64.
b) The probability of rolling all even numbers is also 1/64. There are 3 even numbers on a die (2, 4, and 6), so the probability of rolling an even number on any single roll is 3/6 or 1/2. Since the rolls are independent, the probability of getting an even number on all three rolls is (1/2)(1/2)(1/2) = 1/8. However, we need to subtract the probability of rolling all 4's from this, which is also 1/64. So the final probability is 1/8 - 1/64 = 7/64.
c) The probability of none of the rolls getting a number divisible by 2 is 1 - (1/2)(1/2)(1/2) = 1 - 1/8 = 7/8. This is the complement of rolling all even numbers.
d) The probability of rolling at least one 2 is 1 - the probability of rolling no 2's. The probability of not rolling a 2 on any single roll is 5/6, so the probability of not rolling a 2 on all three rolls is (5/6)(5/6)(5/6) = 125/216. Therefore, the probability of rolling at least one 2 is 1 - 125/216 = 91/216.
e) The probability of rolling all 2's is 1/64. Therefore, the probability of not rolling all 2's is 1 - 1/64 = 63/64.
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