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Sagot :
You are expected to use a z-score table to find how many standard deviations correspond to having 10% outside the bounds. It is like having 68% within 1σ. The table interpolates.
I'm sorry I looked everywhere and I still cannot find how to calculate with the data provided. I'm new to stats so any help is appreciated, if there is any more information let me know and I could solve it.
- Using the normal distribution and the central limit theorem, it is found that the mean is of 64.2 inches, while the standard deviation is of 2.66 inches.
- By the Empirical Rule, 99.7% of American women aged 20-29 years have heights between 58.88 and 69.52 inches.
In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- It measures how many standard deviations the measure is from the mean.
- After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
10% are less than 60.8 inches tall, which means that when [tex]X = 60.8[/tex], Z has a p-value of 0.1, so Z = -1.28, then:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-1.28 = \frac{60.8 - \mu}{\sigma}[/tex]
[tex]60.8 - \mu = -1.28\sigma[/tex]
[tex]\mu = 60.8 + 1.28\sigma[/tex]
10% are more than 67.6 inches tall, hence, due to the symmetry of the normal distribution, when X = 67.6, Z = 1.28.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.28 = \frac{67.6 - \mu}{\sigma}[/tex]
[tex]67.6 - \mu = 1.28\sigma[/tex]
[tex]\mu = 67.6 - 1.28\sigma[/tex]
Then, equaling both equations:
[tex]60.8 + 1.28\sigma = 67.6 - 1.28\sigma[/tex]
[tex]2.56\sigma = 6.8[/tex]
[tex]\sigma = \frac{6.8}{2.56}[/tex]
[tex]\sigma = 2.66[/tex]
[tex]\mu = 67.6 - 1.28\sigma = 67.6 - 1.28(2.66) = 64.2[/tex]
The mean is of 64.2 inches, while the standard deviation is of 2.66 inches.
By the Empirical Rule, 99.7% of the measures are within 3 standard deviations of the mean.
64.2 - 2(2.66) = 58.88
64.2 + 2(2.66) = 69.52
By the Empirical Rule, 99.7% of American women aged 20-29 years have heights between 58.88 and 69.52 inches.
A similar problem is given at https://brainly.com/question/24663213
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