Question

Sheila's doctor is concerned that she may suffer from gestational diabetes (high blood glucose levels during pregnancy). There is variation both in the actual glucose level and in the blood test that measures the level. A patient is classified as having gestational diabetes if the glucose level is above 140 miligrams per deciliter (mg/dl) one hour after having a sugary drink. Sheila's measured glucose level one hour after the sugary drink varies according to the Normal distribution with μμ = 125 mg/dl and σσ = 20 mg/dl.

(a) If a single glucose measurement is made, what is the probability that Sheila is diagnosed as having gestational diabetes?
(b) If measurements are made on 5 separate days and the mean result is compared with the criterion 140 mg/dl, what is the probability that Sheila is diagnosed as having gestational diabetes?

2. Sheila's measured glucose level one hour after a sugary drink varies according to the Normal distribution with

μμ = 125 mg/dl and σσ = 15 mg/dl. What is the level L such that there is probability only 0.1 that the mean glucose level of 4 test results falls above L?

3. Andrew plans to retire in 36 years. He plans to invest part of his retirement funds in stocks, so he seeks out information on past returns. He learns that over the entire 20th century, the real (that is, adjusted for inflation) annual returns on U.S. common stocks had mean 8.7% and standard deviation 20.2%. The distribution of annual returns on common stocks is roughly symmetric, so the mean return over even a moderate number of years is close to Normal.

(a) What is the probability (assuming that the past pattern of variation continues) that the mean annual return on common stocks over the next 36 years will exceed 10%?
(b) What is the probability that the mean return will be less than 6%?

4.) To estimate the mean height

μμ of male students on your campus, you will measure an SRS of students. You know from government data that heights of young men are approximately Normal with standard deviation about 2.8 inches. You want your sample mean x¯¯¯x¯ to estimate μμwith an error of no more than one-half inch in either direction.

(a) What standard deviation must x¯¯¯x¯ have so that 68% of all samples give an x¯¯¯x¯ within one-half inch of μμ? (Use the 68-95-99.7 rule.)
(b) How large an SRS do you need to reduce the standard deviation of x¯¯¯x¯ to the value you found in part(a)?

Answer 1

Solution

Let X = Shiela’s measured glucose level (in mg/dl) one hour after a sugary drink.

We are given X ~ N(µ, σ²) ………………………………………………………….................………… (1)

Also given, “A patient is classified as having gestational diabetes if the glucose level is above 140 miligrams per deciliter (mg/dl) one hour after having a sugary drink.” ……………………...........…. (2)

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 ………………………...............................(3)

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

X bar ~ N(µ, σ²/n),…………………………………………………………….…….................................(5),

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 }] ………………………………..............................…(6)

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

Standard Normal Tables ………………………………………………………………......................... (7a)

or can be found using Excel Function: NORMSDIST(z) which gives P(Z ≤ z) ……....................…(7b)

Percentage points of N(0, 1) can be found using Excel Function: NORMSINV(Probability) which gives values of t for which P(Z ≤ t) = given probability………………. ………...........................................(7c)

Now to work out the solution,

Q1

We are given µ = 125 and σ = 20 ………………………………………………………………………. (8)

Part (a)

If a single glucose measurement is made, probability that Sheila is diagnosed as having gestational diabetes

= P(X > 140) [vide (2)]

= P[Z > {(140 - 125)/20}] [vide (4) and (8)]

= P(Z > 0.75)

= 1 – 0.7734 [vide (7b)]

= 0.2266 ANSWER

Part (b)

If measurements are made on 5 separate days and the mean result is compared with the criterion 140 mg/dl, probability that Sheila is diagnosed as having gestational diabetes

= P(Xbar₅ > 140)

= P[Z > {√5(140 - 125)/(20}] [vide (6) and (8)]

= P(Z > 1.6771)

= 1 – 0.9532 [vide (7b)]

= 0.0468 ANSWER

Q2

We are given µ = 125 and σ = 20 ……………………………………………………………………. (9)

To find the level L such that there is probability only 0.1 that the mean glucose level of 4 test results falls above L.

To satisfy the given condition, we should have: P(Xbar₄ > L) = 0.1

i.e., vide (6) and (8), P[Z > {√4(L - 125)/(20}] = 0.1

=> vide (7c), {√4(L - 125)/(20} = 1.2815

Or, L = 137.81 ANSWER

DONE.