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 this 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 refund program for 3% underperforming toners. Then what should the threshold be?

Step-by-step explanation:

Since the number of pages that this new toner can print is normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = the number of pages.

µ = mean

σ = standard deviation

From the information given,

µ = 2300 pages

σ = 150 pages

1)

the probability that this toner can print more than 2100 pages is expressed as

P(x > 2100) = 1 - P(x ≤ 2100)

For x = 2100,

z = (2100 - 2300)/150 = - 1.33

Looking at the normal distribution table, the probability corresponding to the z score is 0.092

P(x > 2100) = 1 - 0.092 = 0.908

2) P(x < 2200)

z = (x - µ)/σ/√n

n = 10

z = (2200 - 2300)/150/√10

z = - 100/47.43 = - 2.12

Looking at the normal distribution table, the probability corresponding to the z score is 0.017

P(x < 2200) = 0.017

3) for underperforming toners, the z score corresponding to the probability value of 3%(0.03) is

- 1.88

Therefore,

- 1.88 = (x - 2300)/150

150 × - 1.88 = x - 2300

- 288 = x - 2300

x = - 288 + 2300

x = 2018

The threshold should be

x < 2018 pages

i believe it is c.y = 9x

rise over run. 9 units up over 1 unit right or 9/1