A normal distribution has a mean of µ = 100 with σ = 20. If one score is randomly selected from this distribution, what is the probability that the score will have a value between X = 100 and X = 130?

P(100 ≤ x ≤ 130) = 0.43

Step-by-step explanation:

Since the scores are normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = scores

µ = mean score

σ = standard deviation

From the information given,

µ = 100

σ = 20

We want to find the probability that the scores is between 100 and 130. It is expressed as

P(100 ≤ x ≤ 130)

For x = 100,

z = (100 - 100)/20 = 0

Looking at the normal distribution table, the probability corresponding to the z score is 0.5

For x = 100,

z = (130 - 100)/20 = 1.5

Looking at the normal distribution table, the probability corresponding to the z score is 0.93

Therefore,

P(100 ≤ x ≤ 130) = 0.93 - 0.5 = 0.43

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step-by-step explanation:

the answer is 375 miles

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