Answer to Question #207768 in Statistics and Probability for Mary

Question

1. The average cholesterol content of a 

certain canned goods is 215 milligrams 

and the standard deviation is 15

milligrams. Assume the variable is 

normally distributed.

a) If a canned good is selected, what is 

the probability that the cholesterol 

content will be greater 220

milligrams?

b) If a sample of 25 canned goods is 

selected, what is the probability that 

the mean of the sample will be 

larger than 220 milligrams?

2. The number of driving miles before a 

certain kind of tire begins to show wear 

is on the average, 16,800 miles with a 

standard deviation of 3,300 miles. A 

car rental agency buys 36 of these tires 

for replacement purposes and puts 

each one on a different car.

a) What is the probability that the 36 

tires will average less than 16,000

miles until they begin to show 

wear?

b.What is the probability that the 36 

tiles will average more than 18,000

miles until they begin to show 

wear?

Answer 1

1.

a) Let $X=$ cholesterol content of a canned good: $X\sim N(\mu, \sigma^2).$

Given $\mu=215\ mg, \sigma=15\ mg.$

$P(X>220)=1-P(X\leq220)$

$=1-P(Z\leq\dfrac{220-215}{15})\approx1-P(Z\leq0.3333)$

$\approx0.3694$

b) Let $\bar{X}=$ the mean of the sample: $\bar{X}\sim N(\mu, \sigma^2/n).$

Given $\mu=215\ mg, \sigma=15\ mg, n=25$

$P(\bar{X}>220)=1-P(\bar{X}\leq220)$

$=1-P(Z\leq\dfrac{220-215}{15/\sqrt{25}})\approx1-P(Z\geq1.6667)$

$\approx0.0478$

2.

Let $\bar{X}=$the number of driving miles before a certain kind of tire begins to show wear : $\bar{X}\sim N(\mu, \sigma^2/n).$

Given $\mu=16800\ mi, \sigma=3300\ mi, n=36$

a)

$P(\bar{X}<16000)=P(Z<\dfrac{16000-16800}{3300/\sqrt{36}})$

$\approx P(Z<-1.454545)\approx0.0729$

b)

$P(\bar{X}>18000)=1-P(\bar{X}\leq18000)$

$=1-P(Z\leq\dfrac{18000-16800}{3300/\sqrt{36}})$

$\approx1- P(Z\leq 2.181818)\approx0.0146$