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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.

Verified Answer :

LammettHash

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

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|<2)=\mathbb P(-2

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:

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

\(P(x>250)=P(z>2)= 1-P(z<2) =1-0.9772=0.0228 = 2.28\%\)

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

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