Statistics Question, Read Description

Question

In the country of United States of Heightlandia, the height measurements of ten-year-old children are approximately normally distributed with a mean of 55.7 inches, and standard deviation of 4.8 inches.

A) What is the probability that a randomly chosen child has a height of less than 47.2 inches?

Answer= (Round your answer to 3 decimal places.)

B) What is the probability that a randomly chosen child has a height of more than 63.1 inches?

Answer= (Round your answer to 3 decimal places.)

Answer 1

Normal distribution (the famous "bell curve") has the mean in the center of the curve. 1 SD from the mean includes approximately 34% and 2 SD from the mean includes approximately 48%. We can use this to make a guesstimate for these questions.

(A) 47.2 inches is between 1 and 2 SD LEFT of the mean. That's very approximately 9% probability for ≤ 47.2 (50 - 34 - 7 i.e. about 1.5 SD).

(B) 63.1 inches is between 1 and 2 SD RIGHT of the mean. That's very approximately 9% probability for ≥ 63.1 (50 - 34 - 7 i.e. about 1.5 SD)

To work out the exact values, we need to find an exact number of SDs from the mean, which is known as a Z-Score... AND we have to look up this value in a Z score table.

The formula for Z-score is Z = ^{value - mean} / _SD

(A) Z = ^{value - mean} / _SD

=> Z = ^{47.2 - 55.7} / _4.8 = -8.5 / 4.8 = -1.77 (-ve score to left of mean)

from table this gives 0.0384

hence P(h < 47.2) = 3.84%

(B) Z = ^{value - mean} / _SD

=> Z = ^{63.1 - 55.7} / _4.8 = 7.4 / 4.8 = 1.54 (+ve score to right of mean)

from table this gives 0.0618

hence P(h > 63.1) = 6.18%