Question

In a data set with a normal distribution, the mean is 59 and the standard deviation is 3.

Part A

About what percent of the data lie between 53 and 62?

A. 68%

B. 83.85%

C. 81.5%

D. 50%

Part B

Complete: About 95% of the data lie between

Answer 1

Answer:

a) $P(53<X<62)=P(\frac{53-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{62-\mu}{\sigma})=P(\frac{53-59}{3}<Z<\frac{62-59}{3})=P(-2<z<2)$

$P(-2<z<2)=P(z<2)-P(z<-2)$

$P(-2<z<2)=P(z<2)-P(z<-2)=0.97725-0.02275=0.9545$

b) $a=59 +1.96*3=64.88$

And for the other limit we have:

$a=59 -1.96*3=53.12$

And we will have 95% of the values between (53.12, 64.88)

Step-by-step explanation:

Part a

Let X the random variable of interest of a population, and for this case we know the distribution for X is given by:

$X \sim N(59,3)$  

Where $\mu=59$ and $\sigma=3$

We are interested on this probability :

$P(53<X<62)$

And we can solve this using the z score:

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

Using the last formula we got:

$P(53<X<62)=P(\frac{53-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{62-\mu}{\sigma})=P(\frac{53-59}{3}<Z<\frac{62-59}{3})=P(-2<z<2)$

And we can find this probability with this difference:

$P(-2<z<2)=P(z<2)-P(z<-2)$

And uing the normal standard distribution or excel we got:

$P(-2<z<2)=P(z<2)-P(z<-2)=0.97725-0.02275=0.9545$

Part b

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.025$   (a)

$P(X<a)=0.975$   (b)

As we can see on the figure attached the z value that satisfy the condition with 0.95 of the area on the left and 0.05 of the area on the right it's z=1.96.

If we use condition (b) from previous we have this:

$P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.975$  

$P(z<\frac{a-\mu}{\sigma})=0.975$

But we know which value of z satisfy the previous equation so then we can do this:

$z=1.96<\frac{a-59}{3}$

And if we solve for a we got

$a=59 +1.96*3=64.88$

And for the other limit we have:

$a=59 -1.96*3=53.12$

And we will have 95% of the values between (53.12, 64.88)