Question

The number of books borrowed from a library each week follows a normal distribution. When a sample is taken for several weeks, the mean is found to be 190 and the standard deviation is 30. There is a ___% chance that more than 250 books were borrowed in a week.

Answer 1

$\mathbb P(X>250)=\mathbb P\left(\dfrac{X-190}{30}>\dfrac{250-190}{30}\right)=\mathbb P(Z>2)\approx0.0228$

Or, if you're familiar with the empirical rule and the basic properties of the normal distribution, you can use the fact that approximately $95/%$ of the distribution lies within two standard deviations of the mean, or

$\mathbb P(|Z|$

which means about $5\%$ of the distribution lies outside this interval, or

$\mathbb P((-22))\approx0.05$

Because the distribution is symmetric, you have

$\mathbb P(-22)$
$\implies \mathbb P((-22))=\mathbb P(-22)=2\mathbb P(Z>2)\approx0.05$
$\implies \mathbb P(Z>2)\approx 0.025$

Answer 2

Answer:

There is a 2.28% chance that more than 250 books were borrowed in a week.

Step-by-step explanation:

Given : The number of books borrowed from a library each week follows a normal distribution. When a sample is taken for several weeks, the mean is found to be 190 and the standard deviation is 30.

To find : There is a ___% chance that more than 250 books were borrowed in a week ?

Solution :

Formula to find z-score is

$z=\frac{x-\mu}{\sigma}$

Where, $\mu=190$ is the mean

$\sigma=30$ is the standard deviation

x=250 is the value

Substitute the value in the formula,

$z=\frac{250-190}{30}$

$z=\frac{60}{30}$

$z=2$

According to the normal distribution table,  P(z<2) = 0.9772

Now, The probability that more than 250 books were borrowed in a week will be

$P(x>250)=P(z>2)= 1-P(z$

Therefore, There is a 2.28% chance that more than 250 books were borrowed in a week.