Question

A random sample of 50 employees at a company was obtained. The one-way distance from home to work was recorded for each employee in the sample. Suppose the mean of the sample was 15.2 miles and the standard deviation was 4.1 miles.

a) At least what percent of the distances in this sample should we expect to find within 2.7 standard deviations of the mean? Give your answer to the nearest percent.

In a random sample of 750 toner cartridges, the mean number of pages a toner cartridge can print is 4302 and the standard deviation is 340 pages. Assume the distribution of data is normally distributed.

b) The company that makes the toner cartridges guarantees to replace any cartridge that prints fewer than 3282 pages. Approximately how many of the cartridges in the sample would you expect to be replaced under the guarantee policy?

Answer 1

a) We need , P(15.2-2.7 < X < 15.2+2.7 ) = P(12.5 < X < 17.9 ) = P(X < 17.9) - P(X<12.5)

=P((X - 15.2) / 4.1 < (17.9 -15.2) / 4.1 ) - P((X - 15.2 )/ 4.1 < (12.5 -15.2) / 4.1 )

= P(Z<0.6585) - P(Z<-0.6585)

= 2*P(Z<0.6585) - 1

= 2*0.745 - 1 = 0.49 = 49%

At least 49 percent of the distances in this sample we should expect to find within 2.7 standard deviations of the mean.

b)

P(X< 3282) = P( (X - 4302 )/340 < (3282 - 4302) / 340 )

= P(Z < -3 ) = 0.00135

Now, 0.00135*750 = 1.0125 = 1 (approximately)

Approximately 1 cartridge in the sample you would expect to be replaced under the guarantee policy.