Question

Heights of Females The mean height of an adult female in New York City is estimated to be 63.4 inches with a standard deviation of 3.2 inches. What proportion of the adult females in New York City are (a) less than 63.4 inches tall? (b) between 63 and 65 inches tall? (c) taller than 6 feet? (d) exactly 5 feet tall?

Short answer

(a) 50% (b) 24.1% (c) Almost 0% (d) 0%

Step-by-step solution

Calculate for (a)

Since 63.4 inches is the mean height, exactly 50%(0.5) of females are less than 63.4 inches tall, because the mean divides a normal distribution into two equal parts.

Calculate for (b)

To find this, it is easier to find the cumulative probability of each of 63 and 65 separately and then subtract the two. To do this, calculate the Z scores for 63 and 65 individually and find their respective probabilities. Let's denote Z63 as Z score at height 63 and Z65 as Z score at height 65. Let's assume \(P(Z63)\) is the cumulative probability at height 63 and \(P(Z65)\) is the cumulative probability at height 65. \(Z63 = (63 - 63.4) / 3.2 = -0.125\) and \(Z65 = (65 - 63.4) / 3.2 = 0.5\). Looking up these Z scores in a Z table or using a calculator that can compute the probabilities gives \(P(Z63) = 0.4505\) and \(P(Z65) = 0.6915\). Now, subtract these values to get \(P(63<height<65) = P(Z65) - P(Z63) = 0.6915 - 0.4505 = 0.2410\), or 24.1%.

Calculate for (c)

First, convert 6 feet to inches as the mean and standard deviation is given in inches. 6 feet equals 72 inches. Now, calculate the Z score: \(Z72 = (72 - 63.4) / 3.2 = 2.6875\). Looking up this Z-score in any probability table or using any calculator gives a value very close to 1 because this height is so far to right in the distribution, almost all values lie to the left of it. So, the proportion taller than 6 feet will be \(1 - P(Z72)\) which is very close to zero.

Calculate for (d)

In a continuous distribution, the probability of any one specific point, like exactly 5 feet, having been picked is essentially 0 as there are infinite possible values.