Question

A printer toner company launches a new product. the number of pages that this new toner can print is normally distributed with a mean of 2300 pages and a standard deviation of 150 pages.

1. if we select a toner what is the probability that is toner can print more than 2100 pages

2. if we select 10 toners what is the probability that the average page yield of these 10 toners is lower than 2200 pages

3. The company is planning to have a pre-fund program for 3% under performing toners, then what should the threshold be

Answer 1

Solution :

Given that ,

mean = = 2300

standard deviation = = 150

A) P(x > 2100 ) = 1 - p( x< 2100 )

=1- p P[(x - ) / < (2100-2300) /150 ]

=1- P(z < -1.33)

= 1 - 0.0918 = 0.9082

PROBABILITY = 0.9082

B)

n = 10

= = 2300  

= / n = 150/ 10 = 47.43

P( < 2200 ) = P(( - ) / < (2200 - 2300) /47.43 )

= P(z < -2.11 )

= 0.0174

PROBABILITY = 0.0174

C) 3%

P(Z < Z ) = 0 03

z = -1.88

Using z-score formula,

x = z * +

x = (-1.88)  *150+2300

X = 2018