Question

The average number of miligrams (mg) of cholesterol in a cup of a certain brand of ice cream is 660 mg, the standard deviation is 35 mg. Assume the variable is normally distributed.

A. If a cup of ice cream is selected, what is the probability that the cholesterol content will be more than 670 mg?

B. If a sample of 10 cups of ice cream is selected, what is the probability that the mean of the sample will be larger that 670 mg?

Answer 1

Given $\mu=660\ mg, \sigma=35\ mg, n=10$

$Z=\frac {X-\mu}{\sigma/\sqrt{n}} ​$

A. $P(X>670)=P(Z>\frac{670-660}{35})=P(Z>0.29)=1-P(Z<0.29)=0.3859.$

(We found P using z-score table)

The probability that the cholesterol content will be more than 670 mg is 0.3859.

B.

$$P(X>670)=1-P(X\leq670)$$ $$=1-P(Z\leq\dfrac{670-660}{35/\sqrt{10}})$$ $$\approx1-P(Z\leq0.9035)\approx0.1831$$

(We found P using z-score table)

The probability that the mean of the sample will be larger than 670 mg is 0.1831.