Answer to Question #314998 in Statistics and Probability for zelle

Question

The lifespans of the light bulbs produced by a Filip Company are normally distributed with a mean of 1150 hours and a standard deviation of 175 hours. (Note: For z-scores, round off to two decimal places.)

What percentage of the light bulbs would be expected to last between 1000 hours and 1250 hours?

Answer 1

$P(1000<X<1250)=P(\frac{1000-1150}{175}<Z<\frac{1250-1150}{175})=P(-0.86<Z<0.57)=$

$P(Z<0.57)-P(Z<-0.86)=P(Z<0.57)-1+P(Z<0.86)=0.5208=52.08\%.$

Values for P(Z<0.57) and P(Z<0.86) are from the tables for standard normal distribution.