Given a normal distribution with a mean of 125 and a standard deviation of 14, what percentage of values is within the interval 111 to 139?
A. 32%
B. 50%
C. 68%
D. 95%
E. 99.7%

Respuesta :

Answer:68%

Step-by-step explanation:

just took this test

The percentage of values is within the interval 111 to 139 will be "68%".

Probability

According to the question,

Mean = 125

Standard deviation = 14

Let,

The random variable be "X".

Now, the probability:

→ P(111 < X < 139) = P (111 - 125 < X - 125 < 139 - 125)

                           = P(-14 < X - 125 < 14)

                           = P(-[tex]\frac{14}{14}[/tex] < [tex]\frac{(X-125)}{14}[/tex] < [tex]\frac{14}{14}[/tex])

                           = P(-1 < Z < 1)

When, Distribution function "[tex]\phi[/tex]" now,

= [tex]\phi[/tex](1) - [tex]\phi[/tex](-1)

= [tex]\phi[/tex](1) - 1 + [tex]\phi[/tex](1)

= 2 × [tex]\phi[/tex](1) - 1

= 2 × 0.8413 -1

= 1.6826 - 1

= 0.6826 or,

= 68%

Thus the above answer i.e., "option C" is correct.

Find out more information about probability here:

https://brainly.com/question/24756209