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 %