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

for ex: the reciprocal of 3/4 is 4/3. it is the opposite.

reciprocal definition:

a mathematical expression or function so related to another that their product is one; the quantity obtained by dividing the number one by a given quantity.

let x be the money she spent so that she can earn the free t-shirt.

since, she already spent $112.49

⇒

also, as per the statement: jennifer must spend $250 or more in order to qualify for a free t-shirt at the store.

then,

we have an inequality i.,e

subtract 112.49 from both sides we get;

therefore, $137. 51 more money does jennifer need to spend so that she can earn the free t-shirt