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Refer to Exercise 5.18. Find the probability that

  1. x¯is less than 16.
  2. x¯is greater than 23.
  3. x¯is greater than 25.
  4. x¯falls between 16 and 22.
  5. x¯ is less than 14.

Short Answer

Expert verified

a. Probability that x¯is less than 16 is 0.0228.

b. Probability that x¯is greater than 23 is 0.0668.

c. Probability that x¯is greater than 15 is 0.0062.

d. Probability that x¯falls between 16 and 22 is 0.8185.

e. Probability that x¯is less than 14 is 0.00135.

Step by step solution

01

Given information

A random sample of n=64observations is drawn from a population with μ=20and σ=16.

02

Computing the probability that x¯ is less than 16

a.

According to properties of the Sampling distribution of x¯

μx=μand σx¯=σn

Therefore,

μx¯=20and σx¯=1664i.e.σx¯=2

Now,

P(x¯<16)=Px¯-μσln<16-μσln=Px¯-202<16-202=P(z<-2)

Therefore, from z-score table,

P(x¯<16)=0.0228

Thus, probability that x¯is less than 16 is 0.0228.

03

Computing the probability that x¯ is greater than 23 

b.

According to properties of the Sampling distribution of x¯

μx¯=μandσx¯=σn

Therefore,

μx¯=20and σx¯=1664 i.e. σx¯=2

Now,

P(x>23)=Px-μσln>23-μσln=Px-202>23-202=Pz>1.5

Therefore, from z-score table,

P(x>23)=1-Pz<1.5=1-0.9332=0.0668

I

Thus, probability that xis greater than 23 is 0.0668

04

Computing the probability that x is greater than 25

c.

According to properties of the Sampling distribution of x

μx=μand σx=σn

Therefore,

μx=20 and σx=1664 i.e. σx=2

Now,

P(x>25)=Px-μσln>25-μσln=Px-202>25-202=P(z>2.5)

Therefore, from z-score table,

P(x>25)=1-P(z<2.5)=1-0.9937=0.0062

Thus, probability that xis greater than 15 is 0.0062.

05

Computing the probability that x falls between 16 and 22

d.

According to properties of the Sampling distribution of x

μx=μand σx=σn

Therefore,

μx=20 and σx=1664 i.e. σx=2

Now,

P(16<x<22)=P16-μσln<x-μσln<22-μσln=P16-202<x-202<22-202=P-2<z<1

Therefore, from z-score table,

P16<x<22=P-2<z<1=Pz<1-Pz<-2=0.8413-0.02275P(16<x<22)=0.8185

Thus, probability that xfalls between 16 and 22 is 0.8185.

06

Computing the probability that isxless than 14

e.

According to properties of the Sampling distribution of x

μx=μand σxσn

Therefore,

μx=20and σx=1664i.e. σx=2

Now,

Px<14=Px-μσln<14-μσln=Px-202<14-202=P(z<-3)

Therefore, from z-score table,

P(x<14)=P(z<-3)=0.00135

Thus, probability that xis less than 14 is 0.00135.

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Most popular questions from this chapter

Variable life insurance return rates. Refer to the International Journal of Statistical Distributions (Vol. 1, 2015) study of a variable life insurance policy, Exercise 4.97 (p. 262). Recall that a ratio (x) of the rates of return on the investment for two consecutive years was shown to have a normal distribution, with μ=1.5, σ=0.2. Consider a random sample of 100 variable life insurance policies and letx¯represent the mean ratio for the sample.

a. Find E(x) and interpret its value.

b. Find Var(x).

c. Describe the shape of the sampling distribution ofx¯.

d. Find the z-score for the value x¯=1.52.

e. Find Px¯>1.52

f. Would your answers to parts a–e change if the rates (x) of return on the investment for two consecutive years was not normally distributed? Explain.

Question: The standard deviation (or, as it is usually called, the standard error) of the sampling distribution for the sample mean, x¯ , is equal to the standard deviation of the population from which the sample was selected, divided by the square root of the sample size. That is

σX¯=σn

  1. As the sample size is increased, what happens to the standard error of? Why is this property considered important?
  2. Suppose a sample statistic has a standard error that is not a function of the sample size. In other words, the standard error remains constant as n changes. What would this imply about the statistic as an estimator of a population parameter?
  3. Suppose another unbiased estimator (call it A) of the population mean is a sample statistic with a standard error equal to

σA=σn3

Which of the sample statistics,x¯or A, is preferable as an estimator of the population mean? Why?

  1. Suppose that the population standard deviation σis equal to 10 and that the sample size is 64. Calculate the standard errors of x¯and A. Assuming that the sampling distribution of A is approximately normal, interpret the standard errors. Why is the assumption of (approximate) normality unnecessary for the sampling distribution ofx¯?

Suppose xequals the number of heads observed when asingle coin is tossed; that is, x= 0 or x= 1. The population corresponding to xis the set of 0s and 1s generated when thecoin is tossed repeatedly a large number of times. Supposewe select n= 2 observations from this population. (That is,we toss the coin twice and observe two values of x.)

  1. List the three different samples (combinations of 0s and1s) that could be obtained.
  2. Calculate the value of X¯ffor each of the samples.
  3. Show that the sample proportion of 1s, p^, is equal to X¯.
  4. List the values thatp^can assume, and find the probabilitiesof observing these values.
  5. Construct a graph of the sampling distribution ofp^.

Refer to Exercise 5.3. Assume that a random sample of n = 2 measurements is randomly selected from the population.

a. List the different values that the sample median m may assume and find the probability of each. Then give the sampling distribution of the sample median.

b. Construct a probability histogram for the sampling distribution of the sample median and compare it with the probability histogram for the sample mean (Exercise 5.3, part b).

Question:A random sample of n = 500 observations is selected from a binomial population with p = .35.

a. Give the mean and standard deviation of the (repeated) sampling distribution ofp^the sample proportion of successes for the 500 observations.

b. Describe the shape of the sampling distribution of p^. Does your answer depend on the sample size?

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