Answer:
76
Step-by-step explanation:
Final answer:
The worst exam score when compared to the other students is the one with the lowest z-score. In this case, with a z-score of -0.8, the 78 in Math is the worst score.
Explanation:
To determine which exam score was the worst compared to the other students, we need to calculate the number of standard deviations the student's score falls from the class mean. This calculation is known as a z-score. A z-score allows us to understand the position of a score within the context of a distribution.
For the Math exam:
Z = (Your score - Class mean) / Standard deviation
= (78 - 82) / 5
= -0.8.
For the History exam:
Z = (85 - 87) / 4
= -0.5.
For the English exam:
Z = (80 - 76) / 2.5
= 1.6.
The score with the lowest z-score is the worst because it is the farthest below the mean. Comparing the z-scores, the Math exam has the lowest z-score at -0.8, so the 78 in Math is the worst score when compared to the other students in your classes.
Maria ate 1/3of a pie.Her sister,rebecca,ate 1/5of that.what fraction of the whole pie did rebecca eat?
Answer: the fraction of the whole pie that Rebecca ate is 1/15
Step-by-step explanation:
Let x represent the size or area of the whole pie. Maria ate 1/3 of the pie. This means that the amount of pie that maria ate is 1/3 × x = x/3
Her sister,Rebecca ate 1/5 of that. This means that Rebecca ate 1/5 of the amount that Maria ate. The amount that Rebecca ate will be expressed as 1/5 × x/3 = x/15
To determine the fraction of the whole pie that Rebecca ate, we will divide the amount that Rebecca ate by the size of the whole pie. It becomes
x/15/x = x/15 × 1/x
= 1/15
. In the 1980s, it was generally believed that congenital abnormalities affected about 5% of the nation’s children. Some people believe that the increase in the number of chemicals in the environment has led to an increase in the incidence of abnormalities. A recent study examined 384 children and found that 46 of them showed signs of an abnormality. Is this strong evidence that the risk has increased?
a. Define the parameter and state the hypotheses.
b. Define the sampling distribution (mean and standard deviation).
c. Perform the test and calculate P-value
d. State your conclusion.
e. Explain what the p-value means in this context.
Answer:
Reject null hypothesis
Step-by-step explanation:
a) [tex]H_0: p =0.05\\H_a: p >0.05[/tex]
(Right tailed test at 5% level)
p = risk proportion
Sample proportion = [tex]\frac{46}{384} \\\\=0.1198[/tex]
Std error = [tex]\sqrt{\frac{pq}{n} } \\=\sqrt{\frac{0.05(0.95)}{384} } \\=0.0111[/tex]
p difference = [tex]0.1198-0.05=0.0698[/tex]
b) Sampling proportion is Normal with mean = 0.05 and std error = 0.0698
c) Z = test statistic = p diff/std error
= 6.29
p value <0.05
d) Since p < alpha we reject null hypothesis.
e) The probability that null hypothesis is rejected when true is negligible =p value
Answer:
Step-by-step explanation:
Reject null hypothesis
A machine dispenses water into a glass. Assuming that the amount of water dispensed follows a continuous uniform distribution from 10 ounces to 16 ounces, what is the probability that 13 or more ounces will be dispensed in a given glass? .3333 .5000 .6666 .1666
Answer:
I guess they rounded, but the answer is .5000 or 50% chance
Step-by-step explanation:
Count the total chances. There is a chance it will be 10oz, 11oz, 12oz, 13oz, 14oz, 15oz, or 16oz. The question asked 13 or more. That is 13, 14, 15, and 16.
That is a 4/7 chance or probability it will be more than 13oz. Divide 4 and 7 to get your answer.
0.57 or 57%
In a study of red/green color blindness, 550 men and 2400 women are randomly selected and tested. Among the men, 48 have red/green color blindness. Among the women, 5 have red/green color blindness. Test the claim that men have a higher rate of red/green color blindness than women. Also, construct a 98% confidence interval for the difference between color blindness rates of men and women. Does there appear to be a significant difference?
Answer: 98% confidence interval is (0.06,0.11).
Step-by-step explanation:
Since we have given that
Number of men = 550
Number of women = 2400
Number of men have color blindness = 48
Number of women have color blindness = 5
So, [tex]p_M=\dfrac{48}{550}=0.087\\\\p_F=\dfrac{5}{2400}=0.0021[/tex]
So, at 98% confidence interval, z = 2.326
so, interval would be
[tex](p_M-P_F)\pm z\sqrt{\dfrac{p_M(1-p_M)}{n_M}+\dfrac{p_F(1-p_F)}{n_F}}\\\\=(0.087-0.0021)\pm 2.326\sqrt{\dfrac{0.087\times 0.912}{550}+\dfrac{0.0021\times 0.9979}{2400}}\\\\=0.0849\pm 2.326\times 0.012\\\\=(0.0849-0.027912, 0.0849+0.027912)\\\\=(0.06,0.11)[/tex]
Hence, 98% confidence interval is (0.06,0.11).
No, there does not appears to be a significant difference.
In a sample of 24 spools of wire, the average diameter was found to be 3.16mm with a variance of 0.13. Give a point estimate for the population standard deviation of the diameter of the spools of wire. Round your answer to two decimal places, if necessary.
Answer: 0.36
Step-by-step explanation:
Given : Sample size of spools of wire : n= 24
The sample variance = [tex]s^2=0.13[/tex]
The the sample standard deviation should be the square root of the sample variance.
i.e. Sample standard deviation = [tex]s=\sqrt{0.13}=0.360555127546approx0.36[/tex]
Also, the sample standard deviation(s) is the best estimate of population standard deviation [tex](\sigma)[/tex].
Therefore , a point estimate for the population standard deviation [tex](\sigma)[/tex] of the diameter of the spools of wire. =s= 0.36
Hence, a point estimate for the population standard deviation of the diameter of the spools of wire= 0.36
A music professor says that his school's new test for "creative musical analysis" can't be trusted because the test counts and discordant string of pitches as creative. The music professor is attacking the test's:
A) confidence level.
B) margin of error.
C) validity.
D) reliability.
Answer:
C i think i already answerd this
Step-by-step explanation:
The music professor attacking the test is validity. College courses in music-related subjects, such as voice, instruments, music appreciation, theory, and performance, are taught by music professors.
What is meant by music professor ?College courses in music-related subjects, such as voice, instruments, music appreciation, theory, and performance, are taught by music professors. They might instruct sophisticated, highly specialized courses or wide, basic ones. They design curricula, teach classes, and evaluate students' work.
Other names for music teachers include: music educator, band teacher, choir teacher, piano teacher, violin teacher, guitar teacher, elementary school music teacher, high school music teacher, and vocal teacher.
The highest academic position at a college, university, or postsecondary institution is professor. Professors are well-known, competent academics who are frequently regarded as authorities in their fields of study. In addition to graduate courses, a professor teaches upper-level undergraduate courses.
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McPhee Company manufactures rugs in the cutting and assembly process. Rugs are manufactured in 70-rug batch sizes. The cutting time is 14 minutes per rug. The assembly time is 24 minutes per rug. It takes 18 minutes to move a batch of rugs from cutting to assembly. What is the value-added lead time?
Answer:38 min
Step-by-step explanation:
Given
Cutting time [tex]t_1=14 min[/tex]
Assembly Line time [tex]t_2=24 min[/tex]
[tex]t_3=18[/tex] to move a batch of rugs from cutting to assembly
Value-added refers to a process or phase in a system that turns raw materials or work in progress into much more desirable goods and services for downstream consumers.
thus value added time is [tex]t_1+t_2=14+24 =38 min[/tex]
The advertised claim for batteries for cell phones is set at 48 operating hours with proper charging procedures. A study of 5000 batteries is carried out and 15 stop operating prior to 48 hours.
Do these experimental results support the claim that less than 0.2 percent of the company’s batteries will fail during the advertised time period (assuming proper charging procedures were followed)?
Use a hypothesis testing procedure with α = 0.01. State H0 and H1, test statistic, critical value(s), critical (rejection) region, p-value, and conclusion. Hint: Here, H1 is something against what the company claims.
Final answer:
The experimental results support the claim that less than 0.2 percent of the company’s batteries will fail during the advertised time period.
Explanation:
To test if the experimental results support the claim that less than 0.2 percent of the company’s batteries will fail during the advertised time period, we can conduct a hypothesis test.
H0 (null hypothesis): The proportion of batteries that fail during the advertised time period is equal to or greater than 0.2 percent.
H1 (alternative hypothesis): The proportion of batteries that fail during the advertised time period is less than 0.2 percent.
We can use a one-sample proportion test to compare the proportion of batteries that fail in the sample to the claimed proportion. The test statistic for this hypothesis test is the z-test statistic.
The critical value(s) for a one-sided test with α = 0.01 is z = -2.33.
The critical (rejection) region is z < -2.33.
The p-value is calculated by finding the probability of observing 15 or fewer failures out of 5000 batteries assuming that the null hypothesis is true. Using a normal approximation, we can calculate the p-value as the probability of observing x ≤ 15, where x is the number of failures. We find the p-value is less than 0.01.
Since the p-value is less than the significance level of 0.01, we reject the null hypothesis. We have sufficient evidence to support the claim that less than 0.2 percent of the company’s batteries will fail during the advertised time period.
In a random sample of 27 people, the mean commute time to work was 32.8 minutes and the standard deviation was 7.2 minutes. Assume the population is normally distributed and use a t-distribution to construct a 99 % confidence interval for the population mean mu . What is the margin of error of mu ? Interpret the results.
The confidence interval for the population mean ___ , ___
(Round to one decimal place as needed.)
A) Using the given sample statistics, find the left endpoint of the interval using technology, rounding to one decimal place.
B) Using the given sample statistics, find the right endpoint of the interval using technology, rounding to one deci?mal place.
C) Using the formulas and values for the left and/or rightendpoints, find the margin of error from technology, rounding to two decimal places.
Final answer:
To calculate the 99% confidence interval for the population mean commute time, we use the formula CI = x± t * (s/√n), where x is the sample mean, t is the t-value, s is the sample standard deviation, and n is the sample size. By substituting the given values into the formula, we find that the confidence interval is approximately (29.4, 36.2) and the margin of error is 1.7. This means we are 99% confident that the population mean commute time lies within the interval, and the margin of error represents the possible deviation of the sample mean from the population mean.
Explanation:
To construct a 99% confidence interval for the population mean μ, we can use the formula:
CI = x± t * (s/√n)
Where CI is the confidence interval, x is the sample mean, t is the t-value, s is the sample standard deviation, and n is the sample size. In this case, x = 32.8, s = 7.2, and n = 27.
To find the t-value, we need to determine the degrees of freedom (df), which is equal to n - 1. So, df = 27 - 1 = 26. Looking up the t-value for a 99% confidence level with 26 degrees of freedom in a t-table or using calculator software, we find it to be approximately 2.787.
Substituting these values into the formula, we get:
CI = 32.8 ± 2.787 * (7.2/√27)
Calculating this, we find that the confidence interval is approximately (29.4, 36.2).
The margin of error, also known as the error bound for a population mean (EBM), is half the width of the confidence interval. So, the margin of error in this case is (36.2 - 32.8)/2 = 1.7.
Interpreting the results, we can say that we are 99% confident that the population mean μ lies within the range of 29.4 to 36.2 minutes. Furthermore, the margin of error of 1.7 indicates how much the sample mean could vary from the population mean, taking into account the variability in the data.
Consider the sequence −8, −4, 0, 4, 8, 12, ellipsis. Select True or False for each statement. A recursive rule for the sequence is f(1) = −8; f(n) = −4 (n − 1) for all n ≥ 2. A True B False An explicit rule for the sequence is f(n) = −8 + 4 (n − 1). A True B False The tenth term is 28.
Answer:
A recursive rule for the sequence is f(1) = -8; f(n) = -4 (n – 1) for all n ≥ 2 is "FALSE"
An explicit rule for the sequence is f(n) = -8 + 4 (n – 1) is "TRUE"
The tenth term is 28 is "TRUE"
Step-by-step explanation:
Statement (1)
While the first part [f(1) = –8] is TRUE, the second part [f(n) = –4 (n – 1) for all n ≥ 2] would only be true if the sequence ends at the second term.
Check: Since the fifth term of the sequence is 8, then f(5) = 8
From the statement,
f(5) = –4 (5 – 1)
f(5) = –4 × 4 = –16
:. f(5) ≠ 8
Statement (2)
f(n) = –8 + 4 (n – 1) is TRUE
Check: The fifth term of the sequence is 8 [f(5) = 8]
From the statement,
f(5) = –8 + 4 (5 – 1)
f(5) = –8 + 4 (4)
f(5) = –8 + 16 = 8
:. f(5) = 8
Statement (3)
f(10) = 28 is TRUE
Since the explicit rule is TRUE, use to confirm if f(10) = 28:
f(10) = –8 + 4 (10 – 1)
f(10) = –8 + 4 (9)
f(10) = –8 + 36
f(10) = 28
:. f(10) = 28
An article describes a factorial experiment designed to determine factors in a high-energy electron beam process that affect hardness in metals. The first factor is the travel speed in mm/s (three levels: 10, 20, 30). The second fact g
The subject of this question is Physics and the grade level is High School. The article describes a factorial experiment designed to determine factors in a high-energy electron beam process that affect hardness in metals. The experiment focuses on two factors: travel speed and beam current. The aim of the experiment is to determine how these factors affect hardness in metals.
Explanation:In the given article, the subject of the question is physics, specifically the experimentation involving a high-energy electron beam process that affects hardness in metals. The experiment focuses on two factors: travel speed and beam current. The travel speed is measured in mm/s and has three levels: 10, 20, and 30. The beam current is a continuous variable that can be adjusted and is a factor of two. The experiment aims to determine how these factors affect hardness in metals.
A technician compares repair costs for two types of microwave ovens (type I and type II). He believes that the repair cost for type I ovens is greater than the repair cost for type II ovens. A sample of 60 type I ovens has a mean repair cost of $85.79, with a standard deviation of $15.13. A sample of 56 type II ovens has a mean repair cost of $78.67, with a standard deviation of $17.84. Conduct a hypothesis test of the technician's claim at the 0.1 level of significance. Let μ1 be the true mean repair cost for type I ovens and μ2 be the true mean repair cost for type II ovens.Step 1 of 4: State the null and alternative hypotheses for the test.Step 2 of 4: Compute the value of the test statistic. Round your answer to two decimal places.Step 3 of 4: Determine the decision rule for rejecting the null hypothesis H0. Round the numerical portion of your answer to three decimal places.Step 4 of 4: Make the decision for the hypothesis test. Reject or Fail to Reject Null Hypothesis
Answer:
Since p value <0.1 accept the claim that oven I repair costs are more
Step-by-step explanation:
The data given for two types of ovens are summarised below:
Group Group One Group Two
Mean 85.7900 78.6700
SD 15.1300 17.8400
SEM 1.9533 2.3840
N 60 56
Alpha = 10%
[tex]H_0: \mu_1 - \mu_2 =0\\H_a: \mu_1 - \mu_2> 0[/tex]
(Right tailed test)
The mean of Group One minus Group Two equals 7.1200
df = 114
standard error of difference = 3.065
t = 2.3234
p value = 0.0219
If p value <0.10 reject null hypothesis
4) Since p value <0.1 accept the claim that oven I repair costs are more
2log4x(x^3)=5log2x(x)
Good evening ,
Answer:
x=2
Step-by-step explanation:
Look at the photo below for the details.
:)
11. Jillian needs to make a group of 5 people. She has 10 Democrats, 13 Republicans, and 7 Independents to choose from. (a) What is the probability that the group of 5 people will have 2 Democrats, 2 Republicans, and 1 Independent
Answer:
0.1724 or 17.24%
Step-by-step explanation:
There are
10+13+7 = 30 people
so there are C(30,5) combinations of 30 elements taken 5 at a time ways to select groups of 5 people.
[tex]C(30,5)=\displaystyle\binom{30}{5}=\frac{30!}{5!25!}=142,506[/tex]
There are C(10,2) combinations of 10 elements taken 2 at a time ways to select the 2 Democrats
[tex]C(10,2)=\displaystyle\binom{10}{2}=\frac{10!}{2!8!}=45[/tex]
There are C(13,2) combinations of 13 elements taken 2 at a time ways to select the 2 Republicans
[tex]C(13,2)=\displaystyle\binom{13}{2}=\frac{13!}{2!11!}=78[/tex]
There are C(7,1) combinations of 7 elements taken 1 at a time ways to select the Independent
[tex]C(7,1)=\displaystyle\binom{7}{1}=\frac{7!}{1!6!}=7[/tex]
By the Fundamental Principle of Counting, there are
45*78*7 = 24,570
ways of making the group of 5 people with 2 Democrats, 2 Republicans and 1 independent, and the probability of forming the group is
[tex]\displaystyle\frac{24,570}{142,506}=0.1724[/tex] or 17.24%
The Ball Corporation's beverage can manufacturing plant in Fort Atkinson, Wisconsin, uses a metal supplier that provides metal with a known thickness standard deviation σ = .000507 mm. Assume a random sample of 53 sheets of metal resulted in an x⎯⎯ = .3333 mm. Calculate the 90 percent confidence interval for the true mean metal thickness. (Round your answers to 4 decimal places.) The 90% confidence interval is from to
Answer: (0.3332, 0.33341)
Step-by-step explanation:
Formula to find the confidence interval for population mean[tex](\mu)[/tex] :
[tex]\overline{x}\pm z^*\dfrac{\sigma}{\sqrt{n}}[/tex]
, where n= Sample size
[tex]\overline{x}[/tex] = sample mean.
[tex]z^*[/tex] = Critical z-value (two-tailed)
[tex]\sigma[/tex] = population standard deviation.
As per given , we have
n= 53
[tex]\sigma=0.000507\ mm [/tex]
[tex]\overline{x}=0.3333\ mm[/tex]
The critical values for 90% confidence interval : [tex]z^*=\pm1.645[/tex]
Now , the 90 percent confidence interval for the true mean metal thickness:
[tex]0.3333\pm (1.645)\dfrac{0.000507}{\sqrt{53}}\\\\=0.3333\pm(1.645)(0.0000696)\approx0.3333\pm0.00011456\\\\=(0.3333-0.00011456,\ 0.3333+0.00011456)\\\\=(0.33318544,\ 0.33341456)\approx(0.3332,\ 0.3334)[/tex]
Hence, the 90 percent confidence interval for the true mean metal thickness. : (0.3332, 0.3334)
A statistician uses Chebyshev's Theorem to estimate that at least 15 % of a population lies between the values 9 and 20. Use this information to find the values of the population mean, μ , and the population standard deviation σ.
Answer:
\mu = 14.5\\
\sigma = 5.071\\
k = 1.084
Step-by-step explanation:
given that a statistician uses Chebyshev's Theorem to estimate that at least 15 % of a population lies between the values 9 and 20.
i.e. his findings with respect to probability are
[tex]P(9<x<20) \geq 0.15\\P(|x-14.5|<5.5) \geq 0.15[/tex]
Recall Chebyshev's inequality that
[tex]P(|X-\mu |\geq k\sigma )\leq {\frac {1}{k^{2}}}\\P(|X-\mu |\leq k\sigma )\geq 1-{\frac {1}{k^{2}}}\\[/tex]
Comparing with the Ii equation which is appropriate here we find that
[tex]\mu =14.5[/tex]
Next what we find is
[tex]k\sigma = 5.5\\1-\frac{1}{k^2} =0.15\\\frac{1}{k^2}=0.85\\k=1.084\\1.084 (\sigma) = 5.5\\\sigma = 5.071[/tex]
Thus from the given information we find that
[tex]\mu = 14.5\\\sigma = 5.071\\k = 1.084[/tex]
Personnel selection. Suppose that 7 female and 5 male applicants have been successfully screened for 5 positions. If the 5 positions are filled at random from the 12 âfinalists, what is the probability of selecting â
(A) 3 females and 2â males? â
(B) 4 females and 1â male? â
(C) 5â females?
â(D) At least 4â females?
The question involves calculating the probabilities of different combinations of females and males being selected for positions using the hypergeometric distribution, a concept in mathematics related to probability without replacement from two distinct groups.
Explanation:The question is related to the concept of probability in mathematics, specifically the use of the hypergeometric distribution, which is appropriate when sampling without replacement from a finite population divided into two groups. To calculate the probability of various combinations of female and male applicants selected for the positions, we need to use the hypergeometric probability formula:
For part (A), calculating the probability of selecting 3 females and 2 males:For part (B), calculating the probability of selecting 4 females and 1 male:For part (C), calculating the probability of selecting 5 females:For part (D), calculating the probability of selecting at least 4 females (which includes scenarios of selecting 4 or 5 females):These calculations involve combinations (denoted as C(n, k)) and the formula for hypergeometric probability, which is:
Hypergeometric Probability = [C(group1 size, group1 subset size) * C(group2 size, group2 subset size)] / C(total population size, total subset size)
An appliance dealer sells three different models of upright freezers having 14.5, 16.9, and 19.1 cubic feet of storage space. Let x = the amount of storage space purchased by the next customer to buy a freezer. Suppose that x has the following probability distribution.x p(x)14.5 .216.9 .519.1 .3(a) Calculate the mean and standard deviation of x.(b) If the price of the freezer depends on the size of the storage space, x, such that Price = 25x - 8.5, what is the mean value of the variable Price paid by the next customer?(c) What is the standard deviation of the price paid?
The question is about calculating the mean and standard deviation of a discrete random variable representing the size of the freezer model a customer buys, and then the mean and standard deviation of the price paid, which is a function of the freezer size.
Explanation:The random variable X in this scenario represents the size of the freezer model (in cubic feet) that the next customer buys from the appliance dealer. This is a discrete random variable, with possible values being the sizes of the three available freezer models: 14.5, 16.9, and 19.1 cubic feet.
To calculate the mean and standard deviation of X, we can use the formula for the mean of a discrete random variable and the formula for the standard deviation respectively, which are given as follows:
Mean (expected value), E(X) = Σ [x * p(x)]
Standard Deviation, σ(X) = sqrt{Σ [(x - μ)² * p(x)]}
The mean value of the second variable (Price) can be calculated by plugging the expected value of X into the given Price = 25X - 8.5 equation, and the standard deviation by applying the transformation rule for standard deviations (σ_Y = |b| * σ_X, where b is the coefficient of X in the original expression for Y, which in this case is 25).
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: Planning a summer study schedule About Susan plans on studying four different subjects (Math, Science, French, and Social Studies) over the course of the summer. There are 100 days in her summer break and each day she will study one of the four subjects. A schedule consists of a plan for which subject she will study on each day (a) How many ways are there for her to plan her schedule if there are no restrictions on the number of days she studies each of the four subjects? (b) How many ways are there for her to plan her schedule if she decides that the number of days she studies each subject will be the same?
Answer:
Part (A): The total number of ways are [tex]4^{100}[/tex]
Part (B): The total number of ways are [tex]1.6122075076\times10^{57}[/tex]
Step-by-step explanation:
Consider the provided information.
Part(A) a) How many ways are there for her to plan her schedule if there are no restrictions on the number of days she studies each of the four subjects?
She plans on studying four different subjects.
That means she has 4 choices for each day also there is no restriction on the number of days.
Hence, the total number of ways are:
[tex]4\times 4\times4\times4\times......4=4^{100}[/tex]
Part (B) How many ways are there for her to plan her schedule if she decides that the number of days she studies each subject will be the same?
She wants that the number of day she studies each subject will be the same that means she studies for 25 days each subject.
Therefore, the number of ways are:
[tex]^{100}C_{25}\times ^{75}C_{25}\times ^{50}C_{25}\times ^{25}C_{25}=1.6122075076\times10^{57}[/tex]
In the case of no restrictions, there are 4^100 possible study schedules. If each subject is studied the same number of days, the number of possible schedules is found using a permutation with repetitions of multiset formula, resulting in 100! / (25! * 25! * 25! * 25!).
Explanation:This question focuses on combinatorics which is a branch of mathematics that deals with combinations of objects belonging to a finite set. It features the tasks of counting different objects as per certain restrictions, arrangement of objects with regard to considered conditions, and selection of objects satisfying specific criteria.
For part (a), if there are no restrictions, then she could pick any of the four subjects to study each day. For each day, there are 4 choices, so for 100 days, that results in 4^100 possible schedules.
For part (b), if she wants to study each subject the same number of days, then each subject would be studied for 100/4=25 days. This situation is a permutation with repetitions of multiset, which can be found through the formula n!/(r1! * r2! * ... * rk!) where n is total number of items, and r1, r2, ..., rk are numbers of same elements of multiset. Using this equation, the number of ways would be 100! / (25! * 25! * 25! * 25!).
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what is the slope of the line that passes through the points (-2, 5) and (1, 4)
Answer:
-1/3
Step-by-step explanation:
rise / run = slope
4-5/ 1-(-2)
= -1/3
Suppose Clay, a grocery store owner, is monitoring the rate at which customers enter his store. After watching customers enter for several weeks, he determines that the amount of time in between customer arrivals follows an exponential distribution with mean 25
What is the 70th percentile for the amount of time between customers entering Clay's store? Round your answer to the nearest two decimal places. 70th percentile
Answer:
70th percentile for the amount of time between customers entering is 30.10
Step-by-step explanation:
given data
mean = 25
to find out
What is the 70th percentile for the amount of time between customers entering Clay's store
solution
we know that here mean is
mean = [tex]\frac{1}{\lambda}[/tex] ..................1
so here 25 = [tex]\frac{1}{\lambda}[/tex]
and we consider time value corresponding to 70th percentile = x
so we can say
P(X < x) = 1 - [tex]e^{- \lambda *x}[/tex]
P(X < x) = 70 %
1 - [tex]e^{\frac{-x}{25}}[/tex] = 70 %
1 - [tex]e^{\frac{-x}{25} }[/tex] = 0.70
[tex]e^{\frac{- x}{25} }[/tex] = 0.30
take ln both side
[tex]\frac{-x}{25}[/tex] = ln 0.30
[tex]\frac{x}{25}[/tex] = 1.203973
x = 30.10
70th percentile for the amount of time between customers entering is 30.10
The 70th percentile for the amount of time between customers entering Clay's store is 30
How to determine the 70th percentile?The mean of an exponential distribution is:
[tex]E(x) = \frac 1\lambda[/tex]
The mean is 25.
So, we have:
[tex]\frac 1\lambda = 25[/tex]
Solve for [tex]\lambda[/tex]
[tex]\lambda = \frac 1{25}[/tex]
[tex]\lambda = 0.04[/tex]
An exponential function is represented as:
[tex]P(x < x) = 1 - e^{-\lambda x}[/tex]
For the 70th percentile, we have:
[tex]1 - e^{-\lambda x} = 0.70[/tex]
This gives
[tex]e^{-\lambda x} = 1 - 0.70[/tex]
[tex]e^{-\lambda x} = 0.30[/tex]
Substitute [tex]\lambda = 0.04[/tex]
[tex]e^{-0.04x} = 0.30[/tex]
Take the natural logarithm of both sides
[tex]-0.04x = \ln(0.30)[/tex]
This gives
[tex]-0.04x = -1.20[/tex]
Divide both sides by -0.04
[tex]x = 30[/tex]
Hence, the 70th percentile for the amount of time between customers entering Clay's store is 30
Read more about probability distribution at:
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An ANOVA procedure for a two-factor factorial experiment produced the following: a = 6, b = 2, r = 2, SSA = 1.05, SSB = 16.67, SSAB = .60, and SST = 94.52. What is the value of the test statistic for determining whether there is a main effect for factor A? a. 2.63 b. .02 c. The test statistic cannot be computed because SSE is not given. d. .03
Answer:
Which is the output of the formula =AND(12>6;6>3;3>9)?
A.
TRUE
B.
FALSE
C.
12
D.
9
The test statistic for the main effect for factor A cannot be computed from the provided information because the mean square for error (MSE) is not given, which is required for the F statistic calculation.
Explanation:The test statistic for determining whether there is a main effect for factor A is calculated as follows.
First, the mean square for factor A (MSA) is found by dividing SSA by its degrees of freedom, which for factor A is a-1. In this case, with a = 6, degrees of freedom for A is 6-1, which is 5.
Therefore, MSA = SSA / dfA = 1.05 / 5 = 0.21.
The mean square for error (MSE) is not provided, which is necessary to calculate the F statistic for factor A.
To obtain the test statistic for factor A, we need to divide MSA by MSE. Without SSE or MSE, the test statistic cannot be computed. Hence, the correct answer is c.
The test statistic cannot be computed because SSE is not given.
Suppose the returns on long-term government bonds are normally distributed. Assume long-term government bonds have a mean return of 6.0 percent and a standard deviation of 9.9 percent. a. What is the approximate probability that your return on these bonds will be less than −3.9 percent in a given year?
Answer:
0.1587
Step-by-step explanation:
Let X be the random variable that represents a return on long-term government bond. We know that X has a mean of 6.0 and a standard deviation of 9.9, in order to compute the approximate probability that your return on these bonds will be less than -3.9 percent in a given year, we should compute the z-score related to -3.9, i.e., (-3.9-6.0)/9.9 = -1. Therefore, we are looking for P(Z < -1) = 0.1587
A group of 24 people have found 7.2 kg of gold. Assuming the gold is divided evenly, how much gold will each one get in grams?
Answer:
Step-by-step explanation:
Convert 7.2 to grams then divide the grams evenlly to the 24 people then you get 300
Help me please and explain
Answer:
its 114
Step-by-step explanation:
180-66=114 so that's it
Some sources report that the weights of full-term newborn babies in a certain town have a mean of 8 pounds and a standard deviation of 0.6 pounds and are normally distributed. a. What is the probability that one newborn baby will have a weight within 0.6 pounds of the meaning dash that is, between 7.4 and 8.6 pounds, or within one standard deviation of the mean
Answer:
0.682 is the probability that one newborn baby will have a weight within one standard deviation of the mean.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 8 pounds
Standard Deviation, σ = 0.6 pounds
We are given that the distribution of weights of full-term newborn babies is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
P(weight between 7.4 and 8.6 pounds)
[tex]P(7.4 \leq x \leq 8.6) = P(\displaystyle\frac{7.4 - 8}{0.6} \leq z \leq \displaystyle\frac{8.6-8}{0.6}) = P(-1 \leq z \leq 1)\\\\= P(z \leq 1) - P(z < -1)\\= 0.841 - 0.159 = 0.682 = 68.2\%[/tex]
[tex]P(7.4 \leq x \leq 8.6) = 6.82\%[/tex]
This could also be found with the empirical formula.
0.682 is the probability that one newborn baby will have a weight within 0.6 pounds of the mean.
Which of the following expression is equivalent to 6^−7? A 1/(−6)^−7 B 1/6^7 C (−6^)−7 D 1/6^−7
Answer:B
Step-by-step explanation: The answer is B. 1/6^7. Your trying to find the reciprocal of 6^-7. The way you do that is put a 1 over the 6 with a power of -7 like this (6^-7)/(1) and that's how you get your answer!
[tex]\huge\text{Hey there!}[/tex]
[tex]\huge\textbf{This equation should be somewhat}\\\huge\textbf{easy because it has like THREE steps}\\\huge\textbf{to solve it.}[/tex]
[tex]\huge\textbf{But....}[/tex]
[tex]\huge\textbf{Your equation should look like:}[/tex]
[tex]\mathsf{6^{-7}}[/tex]
[tex]\huge\textbf{Now, let's get to the steps:}[/tex]
[tex]\mathsf{6^{-7}}\\\\\mathsf{\approx \dfrac{1}{6\times6\times6\times6\times6\times6\times6}}\\\\\mathsf{\approx \dfrac{1}{6^7}}\\\\\mathsf{\approx \dfrac{1}{279,936}}[/tex]
[tex]\huge\textbf{Therefore, the closet answer to your}\\\huge\textbf{question should be:}[/tex]
[tex]\huge\boxed{\frak{Option\ B. \rm{\dfrac{1}{6^7}}}}\huge\checkmark[/tex]
[tex]\huge\text{Good luck on your assignment \& enjoy your day!}[/tex]
~[tex]\frak{Amphitrite1040:)}[/tex]
Flip a coin 100 times. We want the chance of getting exactly 50 heads. What is the exact probability, correct to six decimal places? What is the Normal approximation of the probability, to six decimal places?
Answer: a) 0.079589 b) 0.079656
Step-by-step explanation:
Since we have given that
Number of times a coin is flipped = 100 times
Number of times he get exactly head = 50
Probability of getting head = [tex]\dfrac{1}{2}[/tex]
We will use "Binomial distribution":
Probability would be
[tex]^{100}C_{50}(\dfrac{1}{2})^{50}(\dfrac{1}{2})^50\\\\=0.079589[/tex]
Using "Normal approximation":
n = 100
p = 0.5
So, mean = [tex]np=100\times 0.5=50[/tex]
Standard deviation is given by
[tex]\sqrt{np(1-p)}\\\\=\sqrt{50(1.05)}\\\\=\sqrt{50\times 0.5}\\\\=\sqrt{25}\\\\=5[/tex]
So,
[tex]P(X<x)=P(Z<\dfrac{\bar{x}-\mu}{\sigma})\\\\So, P(X=50)=P(49.5<X<50.5)\\\\=P(\dfrac{49.5-50}{5}<Z<\dfrac{50.5-50}{5})\\\\=P(-0.1<Z<0.1)\\\\=P(Z<0.1)-P(Z<-0.1)\\\\=0.539828-0.460172\\\\=0.079656[/tex]
Hence, a) 0.079589 b) 0.079656
How much TV do college students watch? A survey of 361 students recorded the number of hours of television they watched per week. The sample mean was 6.504 hours with a standard deviation of 5.584. The standard error of the mean was 0.294. Find a 90% confidence interval for the population mean. Round your answer to three decimal places
Answer: 90% confidence interval would be (6.022,6.986).
Step-by-step explanation:
Since we have given that
Number of students = 361
Sample mean = 6.504
Sample standard deviation = 5.584
Standard error of the mean = 0.294
At 90% confidence level,
So, α = 0.01
So, z = 1.64
Margin of error is given by
[tex]z\times \text{Standard error}\\\\=1.64\times 0.294\\\\=0.48216[/tex]
So, Lower limit would be
[tex]\bar{x}-0.482\\\\=6.504-0.482\\\\=6.022[/tex]
Upper limit would be
[tex]\bar{x}+0.482\\\\=6.504+0.482\\\\=6.986[/tex]
So, 90% confidence interval would be (6.022,6.986).
The recommended daily dietary allowance for zinc among males older than age 50 years is 15 mg/day. An article reports the following summary data on intake for a sample of males age 65−74 years: n = 114, x = 11.2, and s = 6.58. Does this data indicate that average daily zinc intake in the population of all males age 65−74 falls below the recommended allowance? (Use α = 0.05.) State the appropriate null and alternative hypotheses.
Answer:
Step-by-step explanation:
Hello!
To be able to resolve this kind of exercise nice and easy the first step is to determine your study variable and population parameter.
The variable is
X: daily dietary allowance for zinc for a male 50 years old or older. (mg/day)
The parameter of interest is the population mean. (μ)
You need to test if the average zinc intake is bellowed the recommended allowance, symbolically μ < 15. This will be the alternative hypothesis, it's complement will be the null hypothesis. (easy tip to detect the null hypothesis fast, it always carries the = sign)
The hypothesis is:
H₀: μ ≥ 15
H₁: μ < 15
α: 0,05
Now you have no information about the variable distribution. You need the variable to be normally distributed to study the population mean. Since the sample is large enough, you can apply the Central Limit Theorem and approximate the distribution of the sample mean to normal, this way, you can use the approximate Z to make the test.
X[bar]≈N(μ;σ²/n)
Z= X[bar]-μ ≈N(0;1)
√(σ²/n)
Z= 11.2 - 15 = -6.17
6.58/√114
The critical region of this test is one-tailed (left) the critical value is:
[tex]Z_{\alpha } = Z_{0.05} = -1.64[/tex]
If Z ≤ -1.64, you reject the null hypothesis.
If Z > -1.64, you support the null hypothesis
The calculated value is less than the critical value, so the decision is to reject the null hypothesis.
This means that the average daily zinc intake of males age 65 - 74 is below the recommended allowance.
I hope it helps!