Question

The serum cholesterol levels in men aged 18 – 24 are normally distributed with a mean of 178.1 and a standard deviation of 40.7. Units are in mg/100mL. Use R. Paste your commands and output into the answer box.

a) If a man aged 18 – 24 is randomly selected, find the probability that his serum cholesterol level is between 170 and 200.

b) If a sample of 10 men aged 18 – 24 is randomly selected, find the probability that their mean serum cholesterol level is between 180 and 190.

Answer 1

Solution:

Given: The serum cholesterol levels in men aged 18 – 24 are normally distributed with a mean of 178.1 and a standard deviation of 40.7.

Mean = 178.1

Standard Deviation = 40.7

We have to use R to find following probabilities:

Part a) P(  serum cholesterol level is between 170 and 200 ) = ............?

P( 170 < X < 200) =............?

In R we use following command:

pnorm( upper x value , mean=____ , sd=___ ) - pnorm(Lower x value , mean=____ , sd=___ )

thus

pnorm(200, mean=178.1, sd=40.7)-pnorm(170, mean=178.1, sd=40.7)

which gives

0.2836157
thus P( 170 < X < 200) = 0.2836157

( Round answer to specified number of decimal places)

Part b) If a sample of 10 men aged 18 – 24 is randomly selected, find the probability that their mean serum cholesterol level is between 180 and 190.

We have to find:

Since we have to find probability for sample means, we need to find standard error and use this standard error instead of standard deviation to find probability.

Thus use same R command from part a) and use sd = 12.87047

pnorm( upper x value , mean=____ , sd=___ ) - pnorm(Lower x value , mean=____ , sd=___ )

thus

pnorm(190, mean=178.1, sd=12.87047)-pnorm(180, mean=178.1, sd=12.87047)

which gives

0.2637317

that is:

( Round answer to specified number of decimal places)