Question

31. Heights of Women The mean height of women in the United States (ages 20-29) is 64.2 inches. A random sample of 60 women in this age group is selected. What is the probability that the mean height for the sample is greater than 6 Center for Health Statistics) 6 inches? Assume ơ-2.9 inches. (Adapted from National

Answer 1

Solution

Let random variable X represent the height of women, ages 20 -29, of United States.

We are given X has mean µ = 64.2, and σ = 2.9 …………………………………………….. (A)

Back-up Theory

CENTRAL LIMIT THEOREM

Let {X₁, X₂, …, X_n} be a sequence of n independent and identically distributed (i.i.d) random variables drawn from a distribution [i.e., {x₁, x_2, …, x_n} is a random sample of size n] of expected value given by µ and finite variance given by σ². Then, as n gets larger, the distribution of Z = {√n(Xbar − µ)/σ}, approximates the normal distribution with mean 0 and variance 1 (i.e., Standard Normal Distribution). i.e., sample average from any distribution follows Normal Distribution with mean µ and variance σ²/n. ………………………………………………………………..……. (1)

If a random variable X ~ N(µ, σ²), i.e., X has Normal Distribution with mean µ and variance σ²,

then, Z = (X - µ)/σ ~ N(0, 1), i.e., Standard Normal Distribution ……………………………..(2)

X bar ~ N(µ, σ²/n),…………………………………………………………………….....….…….(3)

where X bar is average of a sample of size n from population of X.

So, P(X bar ≤ or ≥ t) = P[Z ≤ or ≥ {(√n)(t - µ)/σ }] …………………………………..…………(4)

Probability values for the Normal Variable, X ~ N(µ, σ²), can be found using Excel Function: Statistical, NORMDIST(x,Mean,Standard_dev,Cumulative), which gives P(X ≤ t) …..........………..(5a)

Probability values for the Standard Normal Variable, Z, can be directly read off from

Standard Normal Tables ………………………………………………………………….…… (6a)

or can be found using Excel Function: Statistical, NORMSDIST .........................………(6b)

Now to work out the solution,

Q31

Vide (A) and (1), the sample mean Xbar is Normal with mean 64.2, and standard deviation = 2.9/√60.

So, probability the mean height of the sample is greater than or equal to 66

= P(Xbar ≥ 66)

= P[Z ≥ {(√60)(66 – 64.2)/2.9 }] [vide (4)]

= P(Z ≥ 4.8076)

= 0.000001 [vide (6b)] Answer

Q33

Probability of randomly selecting a woman with height less than 70 inches

= P(X < 70)

P[Z < {(70 – 64.2)/2.9 }] [vide (2)]

= P(Z < 2.0)

= 0.97725 [vide (6b)]

Probability of randomly selecting a sample of 20 women with mean height less than 70 inches

= P(Xbar < 70)

P[Z < {√20(70 – 64.2)/2.9 }] [vide (4)]

= P(Z < 8.9437)

= 1.0 [vide (6b)]

Since 1.0 > 0.977725,

it is more likely to find mean height of 20 women to be less than 70 inches. Answer

DONE