Question

Suppose Diane and Jack are each attempting to use a simulation to describe the sampling distribution from a population that is skewed right with mean 50 and standard deviation 15. Diane obtains 1000 random samples of size n=3 from the​ population, finds the mean of the​ means, and determines the standard deviation of the means. Jack does the same​ simulation, but obtains 1000 random samples of size n=35 from the population. Complete parts​ (a) through​ (c) below.

​(a) Describe the shape you expect for Diane​'s distribution of sample means. Describe the shape you expect for Jack​'s distribution of sample means. Choose the correct answer below.

A. Diane​'s distribution is expected to be skewed right​, but not as much as the original distribution. Jack​'s distribution is expected to be approximately normal.

B. Diane​'s distribution and Jack​'s distribution are expected to be approximately normal.​ However, Diane​'s will have a greater standard deviation.

C. Jack​'s distribution is expected to be skewed right​, but more than the original distribution. Diane​'s distribution is expected to be approximately normal.

D. Diane​'s distribution is expected to be skewed left​, but not as much as the original distribution. Jack​'s distribution is expected to be approximately normal. ​

(b) What do you expect the mean of Diane​'s distribution to​ be? What do you expect the mean of Jack​'s distribution to​ be? ​

(c) What do you expect the standard deviation of Diane​'s distribution to​ be? What do you expect the standard deviation of Jack​'s distribution to​ be?

Diane​'s distribution is expected to have a standard deviation of

Jack​'s distribution is expected to have a standard deviation of

Answer 1

a) Jack’s distribution is expected to be skewed right, but not as much as the original distribution. Diane’s distribution is expected to be approximately normal.

b) Jack’s distribution is expected to have a mean of 50. Diane’s distribution is expected to have a mean of 50.

c) Diane’s distribution is expected to have a standard deviation of = σ/sqrt (n)

= 15/ sqrt (3)

= 8.66

Jack‘s distribution is expected to have a standard deviation of = σ/sqrt (n)

= 15/ sqrt (35)

= 2.54