Respuesta :
[tex]\mathbb P(X>250)=\mathbb P\left(\dfrac{X-190}{30}>\dfrac{250-190}{30}\right)=\mathbb P(Z>2)\approx0.0228[/tex]
Or, if you're familiar with the empirical rule and the basic properties of the normal distribution, you can use the fact that approximately [tex]95/%[/tex] of the distribution lies within two standard deviations of the mean, or
[tex]\mathbb P(|Z|<2)=\mathbb P(-2<Z<2)\approx 0.95[/tex]
which means about [tex]5\%[/tex] of the distribution lies outside this interval, or
[tex]\mathbb P((-2<Z)\cup(Z>2))\approx0.05[/tex]
Because the distribution is symmetric, you have
[tex]\mathbb P(-2<Z)=\mathbb P(Z>2)[/tex]
[tex]\implies \mathbb P((-2<Z)\cup(Z>2))=\mathbb P(-2<Z)+\mathbb P(Z>2)=2\mathbb P(Z>2)\approx0.05[/tex]
[tex]\implies \mathbb P(Z>2)\approx 0.025[/tex]
Or, if you're familiar with the empirical rule and the basic properties of the normal distribution, you can use the fact that approximately [tex]95/%[/tex] of the distribution lies within two standard deviations of the mean, or
[tex]\mathbb P(|Z|<2)=\mathbb P(-2<Z<2)\approx 0.95[/tex]
which means about [tex]5\%[/tex] of the distribution lies outside this interval, or
[tex]\mathbb P((-2<Z)\cup(Z>2))\approx0.05[/tex]
Because the distribution is symmetric, you have
[tex]\mathbb P(-2<Z)=\mathbb P(Z>2)[/tex]
[tex]\implies \mathbb P((-2<Z)\cup(Z>2))=\mathbb P(-2<Z)+\mathbb P(Z>2)=2\mathbb P(Z>2)\approx0.05[/tex]
[tex]\implies \mathbb P(Z>2)\approx 0.025[/tex]
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
[tex]z=\frac{x-\mu}{\sigma}[/tex]
Where, [tex]\mu=190[/tex] is the mean
[tex]\sigma=30[/tex] is the standard deviation
x=250 is the value
Substitute the value in the formula,
[tex]z=\frac{250-190}{30}[/tex]
[tex]z=\frac{60}{30}[/tex]
[tex]z=2[/tex]
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
[tex]P(x>250)=P(z>2)= 1-P(z<2) =1-0.9772=0.0228 = 2.28\%[/tex]
Therefore, There is a 2.28% chance that more than 250 books were borrowed in a week.
