Solution: We are given that the shoe sizes of American women have a normal distribution, with mean, [tex]\mu=8.47,[/tex] standard deviation [tex]\sigma=1.5[/tex]
We have to find the percentage of American women whose shoe size's are not more than 11.47 P(x<11.47).
We need to first find the z-score.
[tex]z=\frac{x-\mu}{\sigma}[/tex]
[tex]=\frac{11.47-8.47}{1.5}[/tex]
[tex]=\frac{3}{1.5} =2[/tex]
Now we have to find [tex]P(z<2)[/tex]. Using the empirical rule, we know that 97.5% data lies below 2 standard deviations above mean.
Therefore, using the empirical rule, 97.5% of American women have shoe sizes that are no more than 11.47.
The American women have shoe sizes that are no more than 11.47 is 97.5%
The empirical rule states that for a normal distribution, 68% of the distribution are within one standard deviation from the mean, 95% are within two standard deviation from the mean and 99.7% are within three standard deviations from the mean.
Given that:
Mean (μ) = 8.47, Standard deviation (σ) = 1.5
68% are within one standard deviation = μ ± σ = 8.47 ± 1.5 = (6.97, 9.97)
95% are within two standard deviation = μ ± 2σ = 8.47 ± 2 * 1.5 = (5.47, 11.47)
The American women have shoe sizes that are no more than 11.47 = 95% + (100% - 95%)/2 = 97.5%
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