Question

Suppose that shoe sizes of American women have a bell-shaped distribution with a mean of 8.42 and a standard deviation of 1.52. Using the empirical rule, what percentage of American women have shoe sizes that are greater than 9.94? Please do not round your answer.

Answer 1

Answer:

The percentage of the women have size shoes that are greater than

9.94 is 16%

Step-by-step explanation:

* Lets revise the empirical rule

- The Empirical Rule states that almost all data lies within 3  standard

  deviations of the mean for a normal distribution.  

- 68% of the data falls within one standard deviation.  

- 95% of the data lies within two standard deviations.  

- 99.7% of the data lies Within three standard deviations  

* The empirical rule shows that

# 68% falls within the first standard deviation (µ ± σ)

# 95% within the first two standard deviations (µ ± 2σ)

# 99.7% within the first three standard deviations (µ ± 3σ).

* Lets solve the problem

- The shoe sizes of American women have a bell-shaped distribution

  with a mean of 8.42 and a standard deviation of 1.52

∴ μ = 8.42

- The standard deviation is 1.52

∴ σ = 1.52

- One standard deviation (µ ± σ):

∵ (8.42 - 1.52) = 6.9

∵ (8.42 + 1.52) = 9.94

- Two standard deviations (µ ± 2σ):

∵ (8.42 - 2×1.52) = (8.42 - 3.04) = 5.38

∵ (8.42 + 2×1.52) = (8.42 + 3.04) = 11.46

- Three standard deviations (µ ± 3σ):  

∵ (8.42 - 3×1.52) = (8.42 - 4.56) = 3.86

∵ (8.42 + 3×1.52) = (8.42 + 4.56) = 12.98

- We need to find the percent of American women have shoe sizes

  that are greater than 9.94

∵ The empirical rule shows that 68% of the distribution lies  

  within one standard deviation in this case, from 6.9 to 9.94

∵ We need the percentage of greater than 9.94

- That means we need the area under the cure which represents more

  than one standard deviation (more than 68%)

∵ The total area of the curve is 100% and the area within one standard

  deviation is 68%

∴ The area greater than one standard deviation = (100 - 68)/2 = 16

∴ The percentage of the women have size shoes that are greater

   than 9.94 is 16%

Answer 2

You can use the empirical rules for probability related to normal distribution to get the needed percentage.

The percentage of American women having shoe sizes that are greater than 9.94 is 16%

What is empirical rule?

According to the empirical rule, also known as 68-95-99.7 rule, the percentage of values that lie within an interval with 68%, 95% and 99.7% of the values lies within one, two or three standard deviations of the mean of the distribution.

For one standard deviation specifically, we can write it symbolically as

$P(\mu - \sigma < X < \mu + \sigma) = 0.68 = 68\%$

(since probability is count per 1, and percent is count per 100, thus probability times 100 = percent).

where we had $X \sim N(\mu, \sigma)$ (X is normally distributed with mean  $\mu$ and standard deviation  $\sigma$)

Using the above fact to get the needed percentage

It is given that the distribution of shoe size is bell shaped (which is normal distribution).

Let X represents the shoe-size of American women.

Then, we have, by the given data.

$X \sim N(8.42, 1.52)$

Thus, from the empirical rule, for one standard deviation, we get:

$P(\mu - \sigma < X < \mu + \sigma) = 0.68 = 68\%$

$P(6.9 < X < 9.94) = 68\%$

Or

$P(X < 6.9) + P(X > 9.94) = 100 - 68\% = 32\%$

Since normal distribution is symmetric around mean, thus, we have

$P(X < \mu - \sigma) = P(X > \mu + \sigma)\\P(X < 6.9) = P(X > 9.94)$

Thus,

$P(X < 6.9) + P(X > 9.94) = 100 - 68\% = 32\%\\P(X > 9,94) + P(X > 9.94) = 100 - 68\% = 32\%\\\\P(X >9.94) = \dfrac{32}{2}\% = 16\%$

Thus,

The percentage of American women having shoe sizes that are greater than 9.94 is 16%

Learn more about empirical rule here:

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