Question

Suppose x has a distribution with μ = 22 and σ = 20. (a) If a random sample of size n = 37 is drawn, find μx, σ x and P(22 ≤ x ≤ 24). (Round σx to two decimal places and the probability to four decimal places.) μx = σ x = P(22 ≤ x ≤ 24) = (b) If a random sample of size n = 59 is drawn, find μx, σ x and P(22 ≤ x ≤ 24). (Round σ x to two decimal places and the probability to four decimal places.) μx = σ x = P(22 ≤ x ≤ 24) = (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 μx is .

Answer 1

Solution :

Given that,

mean = = 22

standard deviation = = 20

a) n = 37

= = 22

= / n = 20/ 37 = 3.29

P( 22     24 )  

= P[(22 -22) /3.29   ( - ) /    (24 -22) /3.29 )]

= P( 0    Z 0.61 )

= P(Z 0.61 ) - P(Z 0 )

= 0.7291 - 0.50 = 0.2291

probability = 0.2291  

b)

n = 59

= = 22

= / n = 20/ 59 = 2.61

P( 22     24 )  

= P[(22 -22) /2.61   ( - ) /    (24 -22) /2.61 )]

= P( 0    Z 0.0.77 )

= P(Z 0.77 ) - P(Z 0 )

= 0.7794 - 0.50 = 0.2794

probability = 0.2794

c)  The probability of part (b) to be higher than that of part (a).

The standard deviation of part (b)2.61  and part (a)3.29 because of the sample size increased.

The standard deviations are decreasing with increasing sample size.so the probability of part b is higher than part a.