Question

Suppose x has a distribution with μ = 19 and σ = 18. (a) If a random sample of size n = 46 is drawn, find μx, σ x and P(19 ≤ x ≤ 21). (Round σx to two decimal places and the probability to four decimal places.) μx = σ x = P(19 ≤ x ≤ 21) = (b) If a random sample of size n = 64 is drawn, find μx, σ x and P(19 ≤ x ≤ 21). (Round σ x to two decimal places and the probability to four decimal places.) μx = σ x = P(19 ≤ x ≤ 21) = (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 . Need Help? Read It

Answer 1

Solution :

Given that,

mean = = 19

standard deviation = = 18

a) n = 46

=   = 19

= / n = 18 / 46 = 2.65

P( 19 21)  

= P[(19 - 19) /2.65 ( - ) / (21 - 19) /2.65 )]

= P( 0.00 Z 0.75)

= P(Z 0.75 ) - P(Z 0.00 )

Using z table,  

= 0.7734 - 0.5

= 0.2734

b) n = 64

=   = 19

= / n = 18 / 64 = 2.25

P( 19 21)  

= P[(19 - 19) /2.25 ( - ) / (21 - 19) /2.25 )]

= P( 0.00 Z 0.89)

= P(Z 0.89 ) - P(Z 0.00 )

Using z table,  

= 0.8133 - 0.5

= 0.3133

c) part (b) probability is higher because standard deviation is smaller than part ( a) standard deviation