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Heights of females are normally distributed with a mean of 64 inches and a standard deviation of 4 inches. Find the percentage of females that are:

a) less than 59 inches tall;
b) greater than 70 inches tall;
c) between 59 and 70 inches tall.

Answer :

To solve this problem, we'll use the properties of the normal distribution, which is a common continuous probability distribution. Heights of females are normally distributed, meaning they follow a bell-shaped curve centered around the mean. In this case, the mean ([tex]\mu[/tex]) is 64 inches, and the standard deviation ([tex]\sigma[/tex]) is 4 inches.

Standard Normal Distribution (Z-scores)

To find the percentage of females meeting certain height criteria, we need to convert the given heights into Z-scores using the formula:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

where [tex]X[/tex] is the height of interest, [tex]\mu[/tex] is the mean height, and [tex]\sigma[/tex] is the standard deviation.

a) Percentage of females less than 59 inches tall:

  1. Calculate the Z-score for 59 inches:
    [tex]Z = \frac{59 - 64}{4} = \frac{-5}{4} = -1.25[/tex]

  2. Using Z-tables or a standard normal distribution calculator, find the percentage of females with a Z-score less than -1.25.

    • This corresponds to approximately 10.56%.

b) Percentage of females greater than 70 inches tall:

  1. Calculate the Z-score for 70 inches:
    [tex]Z = \frac{70 - 64}{4} = \frac{6}{4} = 1.5[/tex]

  2. Using Z-tables or a standard normal distribution calculator, find the percentage of females with a Z-score greater than 1.5.

    • This corresponds to approximately 6.68%.

c) Percentage of females between 59 and 70 inches tall:

  1. We've already calculated the Z-scores for both 59 inches (-1.25) and 70 inches (1.5).

  2. To find the percentage between these two Z-scores, find the percentage of females less than 1.5 and then subtract the percentage of females less than -1.25.

  3. From Z-tables:

    • Percentage for Z < 1.5 is approximately 93.32%.
    • Percentage for Z < -1.25 is approximately 10.56%.

  4. Subtract these percentages:

    • Percentage between these heights is approximately 93.32% - 10.56% = 82.76%.

These calculations help you understand how data that follows a normal distribution can be analyzed using Z-scores and probability tables to find specific probabilities.

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