Question

For adult men, total cholesterol has a mean of 188 mg/dL and a standard deviation of 43 mg/dL. For adult women, total cholesterol has a mean of 193 mg/dL and a standard deviation of 42 mg/dL. The CDC defines “high cholesterol” as having total cholesterol of 240 mg/dL or higher, “borderline high” as having a total cholesterol of more than 200 but less than 240, and “healthy” as having total cholesterol of 200 or less. A study published in 2017 indicated that about 11.3% of adult men and 13.2% of adult women have high cholesterol. USE EXCEL.

1. An adult woman is randomly selected and her total cholesterol is measured. What is the probability it is in the “borderline high” range?

2. In a study of 45 randomly selected adult men, the number who have high cholesterol is counted. (Assume that 11.3% of men have high cholesterol.)

3. How many of these 45 men do you expect to have high cholesterol?

4. What is the standard deviation for the number of these 45 men that have high cholesterol?

5. What is the probability that at least 10 of these 45 men will have high cholesterol?

Answer 1

Solution

Back-up Theory

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 and hence

P(X ≤ or ≥ t) = P[{(X - µ)/σ} ≤ or ≥ {(t - µ)/σ}] = P[Z ≤ or ≥ {(t - µ)/σ}] .……….........................………...…(1)

Probability values for N(0, 1) can be directly read off from Standard Normal Tables……...............…. (2a)

or can be found using Excel Function: Statistical, NORMSDIST(z) which gives P(Z ≤ z) ................…(2b)

If X ~ B(n, p). i.e., X has Binomial Distribution with parameters n and p, where

n = number of trials and p = probability of one success, then, probability mass function (pmf)

of X is given by p(x) = P(X = x) = (ⁿC_x)(p^x)(1 - p)^{n – x}, x = 0, 1, 2, ……. , n .........................………..(3a)

[This probability can also be directly obtained using Excel Function: Statistical, BINOMDIST]..........(3b)

Mean (average) of X = E(X) = µ = np…............................................................................................. (3c)

Variance of X = V(X) = σ² = np(1 – p).................................................................................. ………..(3d)

Standard Deviation of X = SD(X) = σ = √{np(1 – p)}....................................…...……………………...(3e)

Now to work out the solution,

Let X = cholesterol measurement (mg/dL) of a adult woman.

We assume X ~ N(µ, σ²), where µ = 193, and σ= 42 (given) ........................................................... (4)

Part (1)

Given, “borderline high” as having a total cholesterol of more than 200 but less than 240,

Probability an adult woman randomly selected is in the “borderline high” range

= P(200 < X < 240)

= P[{(200 - 193)/42} < Z < {(240 - 193)/42}] [vide (1) and (4)]

= P(0.1667 < Z < 1.1190)

= P(Z < 1.1190) – P(Z < 0.1667)

= 0.8684 – 0.5662 [vide (2b)]

= 0.3022 Answer 1

Part (2 & 3)

Assuming that 11.3% of men have high cholesterol, expected number of men having high cholesterol in a study group of 45 men

= 45 x 0.113

= 5.085

So, 5 men are expected to have high cholesterol. Answer 2

Part (4)

Let Y = number of men having high cholesterol in a study group of 45 men. Then,

Y ~ B(45, 0.113) ……………………………………………………………………......…..................…. (5)

Vide (3e) and (5),

standard deviation for the number of these 45 men that have high cholesterol

= √(45 x 0.113 x 0.887)

= 2.12 Answer 3

Part (5)

Probability that at least 10 of these 45 men will have high cholesterol

= P(Y ≥ 10)

= 1 – P(Y ≤ 9)

= 1 – 0.9899 [vide (3b)]

= 0.0101   Answer 4

DONE