Question

The average of cholestorel content a certain canned goods is 215 miligrams,and the standarddeviation is 15 miligrams.Assume that the variable is normally distributed.If a sample of 25 canned goods is selected,what is the probability that the mean of the sample will be than 220 miligrams?

Answer 1

mean =215

standard deviation=15

n=25

x=220

P(X>220)=1-P(x$\le$220)

Z=$\frac{x-mean}{\frac{s}{\sqrt n}}$

Z=$\frac{220-215}{15/5}$

Z=$\frac{5}{3}$

P(Z$\le 1.67$ )=0.9525

P(X>220)=1-0.9525

P(X>220)=0.0475

The probability that the mean of the sample will be than 220 milligrams is 0.0475.