Question

Suppose x has a distribution with μ = 15 and σ = 9.

(a) If a random sample of size n = 43 is drawn, find μ_x, σ_x and P(15 ≤ x ≤ 17). (Round σ_x to two decimal places and the probability to four decimal places.)

μ_x =

σ_x =

P(15 ≤ x ≤ 17) =

(b) If a random sample of size n = 67 is drawn, find μ_x, σ_x and P(15 ≤ x ≤ 17). (Round σ_x to two decimal places and the probability to four decimal places.)

μ_x =

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 part (b) is

part (a) because of the sample size. Therefore, the distribution about μ_x is .

Answer 1

solution:-

given that μ = 15 and σ = 9

here given n = 43

(a) μx = 15
  
σx = σ/sqrt(n) = 9/sqrt(43) = 1.37

=> P(15 ≤ x ≤ 17) = P((15-15)/1.37 < z < (17-15)/1.37)

= P(0 < z < 1.46)

= P(z < 1.46) - P(z < 0)

= 0.9279 - 0.5

= 0.4279

(b) here given that n = 67

μx = 15
  
σx = σ/sqrt(n) = 9/sqrt(67) = 1.10

=> P(15 ≤ x ≤ 17) = P((15-15)/1.10 < z < (17-15)/1.10)

= P(0 < z < 1.82)

= P(z < 1.82) - P(z < 0)

= 0.9656 - 0.5

= 0.4656

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