In a random sample of 750 toner cartridges, the mean number of pages a toner car

Question

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.

a. Between what two values can you expect to find (the middle) 99.7% of the sample data?

b. What percent of the sample data should you expect to be more than 1 standard deviation away from the mean?

c. How many of the toner cartridges in the sample should you expect to print between 3962 pages and 4642 pages?

d. The company that makes the toner cartridges guarantees to replace any cartridge that prints fewer than 3622 pages. Approximately how many of the cartridges in the sample would you expect to be replaced under the guarantee policy?

Explanation / Answer

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.

a. Between what two values can you expect to find (the middle) 99.7% of the sample data?

99.7% of the sample data falls within 3 sd limits.

Lower limit = mean -3*sd =4302-3*340 = 3282

Upper limit = mean +3*sd =4302+3*340 = 5322

b. What percent of the sample data should you expect to be more than 1 standard deviation away from the mean?

Mean+sd =4302+340=4642

Z value for 4642, z=(4642-4302)/340 =1

P( x >4642) = P( z >1) =0.1587

The required percentage =15.87%

c. How many of the toner cartridges in the sample should you expect to print between 3962 pages and 4642 pages?

Z value for 4642, z=(4642-4302)/340 =1

Z value for 3692, z=(3692-4302)/340 = -1.79

P( 3692<x,4302) = P( -1.79<z<1)

P( z<1) – P( z < -1.79)

= 0.8413 - 0.0367

=0.8046

Expected cartridges =0.8046*750=603.45

=603 ( rounded)

d. The company that makes the toner cartridges guarantees to replace any cartridge that prints fewer than 3622 pages. Approximately how many of the cartridges in the sample would you expect to be replaced under the guarantee policy?

Z value for 3622, z=(3622-4302)/340 =-2

P( x <3622) = P( z < -2) =0.0228

0.0228*750

=17.1

=17 ( rounded)

We expect 17 cartridges to be replaced under the guarantee policy.

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.