Question

Suppose x has a distribution with μ = 14 and σ = 6.

(a) If a random sample of size n = 31 is drawn, find μ_x, σ_x and P(14 ≤ x ≤ 16). (Round σ_x to two decimal places and the probability to four decimal places.)

μx =

σx =

P(14 ≤ x ≤ 16) =

(b) If a random sample of size n = 61 is drawn, find μ_x, σ_x and P(14 ≤ x ≤ 16). (Round σ_x to two decimal places and the probability to four decimal places.)

μx =

σx =

P(14 ≤ x ≤ 16) =

(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--- larger than smaller than the same as part (a) because of the  ---Select--- smaller larger same sample size. Therefore, the distribution about μ_x is  ---Select--- the same narrower wider .

Answer 1

Solution :

Given that,

mean = = 14

standard deviation = = 6

a) n = 31

= = 14

= / n = 6 / 31 = 1.08

P(14 16)  

= P[(14 - 14) / 1.08 ( - ) / (16 - 14) / 1.08 )]

= P( 0 Z 1.85)

= P(Z 1.85) - P(Z 0)

Using z table,  

= 0.9678 - 0.5  

= 0.4678

b) n = 61

= = 14

= / n = 6 / 61 = 0.77

P(14 16)  

= P[(14 - 14) / 0.77   ( - ) / (16 - 14) / 0.77)]

= P( 0 Z 2.60)

= P(Z 2.60) - P(Z 0)

Using z table,  

= 0.9953 - 0.5  

= 0.4953

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