Question

suppose medical records indicate that the length of newborn babies (in inches) is normally distributed with a mean of 20 and a standard deviation of 2.6. find the probability that a given infant is between 17.4 and 22.6 inches long

Answer 1

Answer:

• 0.6826 (68.26%)

Explanation:

1) Find the z-scores:

• z = [X - μ ] / σ

a) z-score for 22.6 inches length

• X = 22.6
• μ = 20
• σ = 2.6

• z = [ 22.6 - 20 ] / 2.6 = 1.00

b) z-score for 17.4 inches length

• X = 17.4
• μ = 20
• σ = 2.6

• z = [ 17.4 - 20 ] / 2.6 =  - 1.00

2) Probability

Then, you have to find the probability that the length of an infant is between - 1.00 and 1.00 standards deviations (σ) from the mean (μ).

That is a well known value of 68%, which is part of the 68-95-99.7 empirical rule.

The most exact result is obtained from tables and is 68.26%:

• 1 - P (z ≥ 1.00) - P (z ≤ - 1.00) = 1 - 0.1587 - 0.1587 = 0.6826 = 68.26%