Assuming a normal distribution with a true mean of 5 inches and a standard deviation of 2.2 inches, what is the probability (in percentage) that future measurements will fall above 4.35 inches?

P = 61.41 %

Explanation:

Let's call X to the random variable.

X ~ N(μ,σ)

Where μ is the mean and σ is the standar deviation.

They are asking us about P(X>4.35 inches)

This probability is equal to :

P(X>4.35 inches) = 1 - P(X≤4.35 inches)

Writing ≤ or < is the same because normal distribution is a continuos random variable.

Let's find first P(X≤4.35 inches). We need to turn X into a N(0,1) standardizing the variable X

We do this by subtracting the mean and dividing by the standard deviation

P(X≤4.35 inches) = P [(X-μ)/σ ≤ (4.35 inches - μ)/σ]

(X-μ)/σ = Z ⇒Z ~ N(0,1)

(4.35 inches - μ)/σ = (4.35 inches-5 inches)/2.2 inches = -0.295454

P(X≤4.35 inches) = P(Z≤-0.295454)

The probability for Z we can find it on a table

P(Z≤-0.295454) =Ф (-0.295454) = 0.3859 = P(X≤4.35 inches)

Then P(X>4.35 inches) = 1 - P(X≤4.35 inches) = 1 - 0.3859 = 0.6141 = 61.41 %

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answer:

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