Question

The manufacturer of a new line of ink-jet printers would like to include as part of its advertising the number of pages a user can expect from a print cartridge. The results from a sample of 14 cartridges can be found in the Excel test data file. Numbers are

1924
1518
1641
2358
1305
1272
2207
1580
1958
1069
1207
1453
1940
1331

1. What are the point estimates for the population mean and population standard deviation?
2. Develop a 95% interval for the population mean.
3. If the point estimate of the population mean and standard deviations are accurate, what is the probability that another sample of 14 cartridges will last an average of more than 1750 pages?

Answer 1

Solution-:

By using R-softyware:

> x=c(1921,1518,1641,2358,1305,1272,2207,1580,1958,1069,1207,1453,1940,1331);x
[1] 1921 1518 1641 2358 1305 1272 2207 1580 1958 1069 1207 1453 1940 1331
> mean=mean(x);mean # Sample mean
[1] 1625.714
> sqr=var(x);sqr # sample variance
[1] 154950.4
> s=sqrt(sqr);s # sample standard devation
[1] 393.6374
> s=sqrt(sqr);s
[1] 393.6374

> n=14
> var=((n-1)/n)*sqr;var # Population variance
[1] 143882.5
> sqrt(var)   # Population standard devation
[1] 379.3185

(a) The point estimates for the population mean is  

And The point estimates for the population population standard deviation is

(b) The 95% C.I. for population mean is,

Here,

Therefore, the required C.I. is

(c) We find  the probability [ another sample of 14 cartridges will last an average of more than 1750 pages]

  

  

From Normal Integral table

Therefore, the probability that another sample of 14 cartridges will last an average of more than 1750 pages is 0.1094