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

Answer:

1. 90.82% probability that the toner can print more than 2100 pages

2. 1.74% probability that the average page yield of these 10 toners is lower than 2200 pages

3. The threshold shoulb be 2018 pages

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

$\mu = 2300, \sigma = 150$

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

This is 1 subtracted by the pvalue of Z when X = 2100. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{2100 - 2300}{150}$

$Z = -1.33$

$Z = -1.33$ has a pvalue of 0.0918

1 - 0.0918 = 0.9082

90.82% probability that the 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

Now we apply the central limit theorem, so $n = 10, s = \frac{150}{\sqrt{10}} = 47.43$

This probability is the pvalue of Z when X = 2200. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{2200 - 2300}{47.43}$

$Z = -2.11$

$Z = -2.11$ has a pvalue of 0.0174

1.74% 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

The threshold should be the 3rd percentile, which is the value of X when Z has a pvalue of 0.03. So it is X when Z = -1.88.

$Z = \frac{X - \mu}{\sigma}$

$-1.88 = \frac{X - 2300}{150}$

$X - 2300 = -1.88*150$

$X = 2018$

The threshold shoulb be 2018 pages