e Chapter 7 Homework A) Amazon com-Online Sh e.CengageNOWu2 Online t TripAdvisor

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e Chapter 7 Homework A) Amazon com-Online Sh e.CengageNOWu2 Online t TripAdvisor Chapter 7 Homework Question 8 (of 10) 8. 00 ponts A random sample is drawn from a population with mean ?-69 and standard deviation ? = 5.8, Use Table a. Is the sampling distribution of the sample mean with n # 17 and n-42 normally distributed? O Yes, both the sample means we have a normal distribution O No, both the sample means will not have a normal distribution O No, only the sample mean with n 17 will have a normal distribution No, only the sample mean with n: 42 will have a normal distribution. b. Can you use the standard normal distribution to calculate the probability that the sample mean falls between 69 and 72 for both sample sizes? O Yes, for both the sample sizes, standard normal distribution could be used O No, for both the sample sizes, standard normal distribution could not be used O No, only for the sample size with n 17, standard normal distribution could be used No, only for the sample size with n 42, standard normal distribution could be used c. Calculate the probability that the sample mean fallis between 69 and 72 for n 42 (Round intermediate calculations to 4 decimal places, "z value to 2 decimal places, and final answer to 4 decimal places.) ?-0.0319. Probability

Explanation / Answer

8.

We have been population distribution params. Its not known at this stage whether they are normal distribution' or not.

Mean = 69
Stdev = 5.8

a. No, not the one with sample size of 17. The one with sample size of n = 42 will be normally distributed

The rule is that only samples which are sufficiently sized ( i.e larger than 30) are normally distributed

b. P(69<X<72)?. To use standard normal distribution we need to have a sample size above 30. Other
use t-distribution. The central limit theorm is what guides this. Here is it' concept - Central Limit Theorem
The central limit theorem states that if you have a population with mean ? and standard deviation ? and take
sufficiently large random samples from the population with replacementtext annotation indicator, then the distribution of the sample means will be approximately normally distributed

c. P(69<X<72)

Using the formula: Z = (X-Mu)/(Sigma/sqrt(n))

= P(69-69/(5.8/sqrt(42)) <Z< (72-69)/(5.8/sqrt(42)))

= P(0<Z<3.35)

Now, use the Z table to get the probability values:

= .9996-.50

= .4996

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