Question

Suppose x has a distribution with μ = 21 and σ = 15.

(a) If a random sample of size n = 36 is drawn, find μ_x, σ_x and P(21 ≤ x ≤ 23). (Round σ_x to two decimal places and the probability to four decimal places.)

μ_x =

σ_x =

P(21 ≤ x ≤ 23) =

(b) If a random sample of size n = 60 is drawn, find μ_x, σ_x and P(21 ≤ x ≤ 23). (Round σ_x to two decimal places and the probability to four decimal places.)

μ_x =

σ_x =

P(21 ≤ x ≤ 23) =

(c) Why should you expect the probability of part (b) to be higher than that of part (a)? (Hint: Consider the standard deviations in parts (a) and (b).)
The standard deviation of part (b) is  ---Select--- smaller than larger than the same as part (a) because of the  ---Select--- smaller same larger sample size. Therefore, the distribution about μ_x is  ---Select--- wider the same narrower .

Answer 1

a)

μ_x = population mean =21

std error=σx̅=σ/√n=15/√36 =

2.5000

probability =P(21<X<23)=P((21-21)/2.5)<Z<(23-21)/2.5)=P(0<Z<0.8)=0.7881-0.5=0.2881

b)

μ_x = population mean =21

std error=σx̅=σ/√n=15/√60 =

1.9365

probability =P(21<X<23)=P((21-21)/1.936)<Z<(23-21)/1.936)=P(0<Z<1.03)=0.8485-0.5=0.3485

c)

The standard deviation of part (b) is smallerthen part (a) because of the larger sample size.

Therefore, the distribution about μ_x is narrower