Question

Consider a large population of interest. It's distribution is normal and it's mean is 176 and standard deviation is 101. Let X = a single observation.

(Round all probabilities to four decimals)

a) Find P(X < 181):

A random sample of 148 is taken. Let X¯¯¯X¯ = sample average of the observations.

b) Find P(X¯¯¯X¯ < 181):

Consider another large population of interest. It's distribution is unknown and it's mean is 176 and it standard deviation is 101. Let Y = a single observation.

A random sample of 148 is taken. Let Y¯¯¯Y¯= sample average of the observations.

c) Find P(Y¯¯¯Y¯ < 181):

Answer 1

(a)

$\mu$ = 176

$\sigma$=101

To find P(X<181):

Z = (181 - 176)/101 = 0.0495

Table of Area Under Standard Normal Curve gives area = 0.0199

So,

P(X<181) = 0.5 + 0.0199 = 0.5199

(b)

$\mu$ = 176

$\sigma$=101

n = 148

SE = $\sigma$ /$\sqrt{n}$

= 101/$\sqrt{148}$ = 8.3021

To find P($\bar{x}$<181):

Z = (181 - 176)/8.3021 = 0.6023

Table of Area Under Standard Normal Curve gives area = 0.2257

So,

P($\bar{x}$<181) = 0.5 + 0.2257 = 0.7257

(c)

By Central Limit Theorem, the sampling distribution of sample means is normal distribution even if the population distribution is not normal distribution.

So,

$\mu$ = 176

$\sigma$=101

n = 148

SE = $\sigma$ /$\sqrt{n}$

= 101/$\sqrt{148}$ = 8.3021

To find P($\bar{x}$<181):

Z = (181 - 176)/8.3021 = 0.6023

Table of Area Under Standard Normal Curve gives area = 0.2257

So,

P($\bar{x}$<181) = 0.5 + 0.2257 = 0.7257