Question

Suppose medical records indicate thatthe length of newborn babies (ininches) is normally distributed with amean of 20 and a standard deviation of2.6. Find the probability that a giveninfant is between 17.4 and 22.6 incheslong

Answer 1

P(17.4 ≤ X ≤ 22.6)

Using the formula:

Z = (X - μ)/σ

Where:

μ = Mean = 20

σ = Standard deviation = 2.6

P(17.4 ≤ X ≤ 22.6) = P ( (X - μ)/σ ≤ Z ≤ (X - μ)/σ) )

Replacing the data:

P ( (X - μ)/σ ≤ Z ≤ (X - μ)/σ) )

P ( (17.4 - 20)/2.6 ≤ Z ≤ (22.6-20)/2.6)

P ( -1 ≤ Z ≤ 1)

Now, let's express P ( -1 ≤ Z ≤ 1) as:

P ( -1 ≤ Z ≤ 1) = P(Z ≤ 1) - (1 - P(Z ≤ 1) )

P(Z ≤ 1) = 0.8413

Therefore:

P(17.4 ≤ X ≤ 22.6) = 0.8413 - (1-0.8413) = 0.8413 - 0.1587 = 0.6826

You can express the result as a percentage:

0.6826*100 = 68.26%