Question

Suppose x has a distribution with μ = 15 and σ = 12.

(a) If a random sample of size n = 32 is drawn, find μx, σ x and P(15 ≤ x ≤ 17). (Round σx to two decimal places and the probability to four decimal places.)

μx =

σ x =

P(15 ≤ x ≤ 17) =

(b) If a random sample of size n = 57 is drawn, find μx, σ x and P(15 ≤ x ≤ 17). (Round σ x to two decimal places and the probability to four decimal places.)

μx =

σ x =

P(15 ≤ x ≤ 17) = (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 is smaller than because of the larger sample size. Therefore, the distribution about μx is wider, the same, narrower

Answer 1

Answer:

Given,

Mean = 15

Standard deviation = 12

Sample n = 32

Standard error = s/sqrt(n)

= 12/sqrt(32)

= 2.12

P(15 <= x <= 17) = P((15 - 15)/2.12 < z < (17 - 15)/2.12))

= P(0 < z < 0.94)

= P(z < 0.94) - P(z < 0)

= 0.8263912 - 0.6914625 [since from z table]

= 0.1349

b)

Sample size n = 57

P(15 <= X <= 17) = P((15 - 15)/(12/sqrt(57)) < z < (17 - 15)/(12/sqrt(57)))

= P(0 < z < 1.26)

= P(z < 1.26) - P(z < 0)

= 0.8961653 - 0.6914625 [since from z table]

= 0.2047

The standard deviation of part (b) is smaller than part (a) because of the larger sample size. Therefore, the distribution about μ_x is the same.