Question

A population of scores forms a normal distribution with a mean of μ = 82 and a standard deviation of σ = 26.

(a) What proportion of the scores in the population have values less than X = 86? (Round your answer to four decimal places.)

(b) If samples of size n = 15 are selected from the population, what proportion of the samples will have means less than M = 86? (Round your answer to four decimal places.)


(c) If samples of size n = 35 are selected from the population, what proportion of the samples will have means less than M = 86? (Round your answer to four decimal places.)

Answer 1

a)

Given,

= 82, = 26

We convert this to standard normal as

P( X < x) = P( Z < x - / )

So,

P( X < 86) = P( Z < 86 - 82 / 26)

= P( Z < 0.1538)

= 0.5611

b)

Using central limit theorem,

P( < x) = P( Z < x - / / sqrt(n) )

P( M < 86) = P( Z < 86 - 82 / 26 / sqrt(15) )

= P( Z < 0.5958)

= 0.7243

c)

P( M < 86) = P( Z < 86 - 82 / 26 / sqrt(35) )

= P( Z < 0.9102)

= 0.8186