Question

Suppose x has a distribution with μ = 23 and σ = 18.

(a) If a random sample of size n = 42 is drawn, find μ_x, σ_x and P(23 ≤ x ≤ 25). (Round σ_x to two decimal places and the probability to four decimal places.)

μx =

σx =

P(23 ≤ x ≤ 25) =

(b) If a random sample of size n = 63 is drawn, find μ_x, σ_x and P(23 ≤ x ≤ 25). (Round σ_x to two decimal places and the probability to four decimal places.)

μx =

σx =

P(23 ≤ x ≤ 25) =

(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--- larger smaller same sample size. Therefore, the distribution about μ_x is  ---Select--- wider narrower the same .

Answer 1

Solution :

Given that,

mean = = 23

standard deviation = = 18

(A) n=42

= 23

=  / n = 18 / 42=2.7775

= P(23<    < 25) = P[(23 - 23) / 2.7775< ( - ) / < (25 - 23) / 2.7775)]

= P( 0< Z <0.72 )

= P(Z <0.72 ) - P(Z <0 )

Using z table,  

=0.7642 -0.5   

= 0.2642

(B)n=63

= 23

=  / n = 18 / 63=2.2678

= P(23<    < 25) = P[(23 - 23) / 2.2678< ( - ) / < (25 - 23) /2.2678)]

= P( 0< Z <0.88)

= P(Z <0.88 ) - P(Z <0)

Using z table,  

=0.8106 -0.5   

= 0.3106