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 = 15 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 7 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?

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 11%?
(b) What is the probability that the mean return will be less than 4%?

Answer 1

1)a) P(X > 140)

     = P((X - )/ > (140 - )/)

     = P(Z > (140 - 125)/15)

     = P(Z > 1)

     = 1 - P(Z < 1)

     = 1 - 0.8413 = 0.1587

b) P( > 140)

= P(( - )/() > (140 - )/() )

= P(Z > (140 - 125)/(15/))

= P(Z > 2.65)

= 1 - P(Z < 2.65)

= 1 - 0.9960

= 0.0040

2)a) P(X > 11)

     = P((X - )/ > (11 - )/)

    = P(Z > (11 - 8.7)/20.2)

    = P(Z > 0.11)

    = 1 - P(Z < 0.11)

    = 1 - 0.5438

    = 0.4562

b) P(X < 4)

     = P((X - )/ < (4 - )/)

    = P(Z < (4 - 8.7)/20.2)

    = P(Z < -0.23)

    = 0.4090