Find the probability that a value randomly selected from this population will have a value greater than 130. b. Find the probability that a value randomly selected from this population will have a value less than 95. c. Find the probability that a value randomly selected from this population will have a value between 95 and 130.

a) 0.0668

b) 0.3085

c) 0.6247

Question:

A population is normally distributed with μ = 100 and σ= 20.

a. Find the probability that a value randomly selected from this population will have a value greater than

130.

b. Find the probability that a value randomly selected from this population will have a value less than 90.

c. Find the probability that a value randomly selected from this population will have a value between 90

and 130.

Step-by-step explanation:

a) The probability that a value randomly selected from this population will have a value greater than 130 = P(X >130)

A z-score also referred to as a standard normal table shows the number of standard deviations a raw score lays either above or below the mean.

Let's determine the z-score using:

z = (x - µ)/σ

µ = 100

σ = 20

z = (130-100)/20

z = 30/20 = 1.5

The probability from the standard normal table associated with z = 1.50 is 0.4332. This is the area between z = 1.50 and the mean.

The total area under any normal curve is 1. Due to the fact that the normal curve is symmetric about the mean, the area on either sides of the mean is 0.5.

The desired probability = 0.5000 - 0.4332

= 0.0668

Find attached the diagram.

b) The probability that a value randomly selected from this population will have a value less than 90 = P(X<90)

Using the z-score formula:

z = (x - µ)/σ

z = (90-100)/20

z = (-10)/20 = -0.5

The probability from the standard normal table associated with

z = -0.50 is 0.1915. This is the area between z = -0.50 and the mean.

The desired probability = 0.5000 - 0.1915

= 0.3085

Find attached the diagram.

c) The probability that a value randomly selected from this population will have a value between 90 and 130 = P(90≤X≤130)

We earlier found the z score when P(X<90) = -0.5

And the probability associated with

z = -0.50 is 0.1915.

The z score when P(X>130) = 1.5

And the probability associated with z = 1.50 is 0.4332.

The desired probability is the addition of the two probabilities:

P(90≤X≤130) = 0.1915+0.4332

P(90≤X≤130) = 0.6247

yes

step-by-step explanation:

30 people weigh 160 pounds or more