Heights of adult women have a mean of 63.6 in. and a standard deviation of 25 in

Question

Heights of adult women have a mean of 63.6 in. and a standard deviation of 25 in. Does Chebyshev's Theorem say about the percentage of women with heights between 58.6 in. and 68.6 in.? At least 75% of the heights should fall between 58.6 in. and 68.6 in. 75% of the heights should fall between 58.6 in. and 68.6 in. At least 89% of the heights should fall between 58.6 in. and 68.6 in. 89% of the heights should fall between 58.6 in. and 68.6 in. Heights of adult women have a mean of 63.6 in. and a standard deviation of 2.5 in. Apply Chebyshev's Theorem to the data using k = 3. Interpret the results. (56.1, 71.1) At least 89% of the heights are between 56.1 and 71.1 inches. (56.1.71.1) 89% of the heights are between 56.1 and 71.1 inches. (56.1.71.1) At least 99% of the heights are between 56.1 and 71.1 inches. (56.1.71.1) 99% of the heights are between 56.1 and 71.1 inches.

Explanation / Answer

According to the Chebyshev's theorem, at least 11/k2 of the data lie within k standard deviations of the mean

Thus, between 68.6 And 58.6, means mean is 68.6-63.6=5 away from the upper limit or it is 5/2.5 =2 standard deviation away. Thus 1-1/(2)2 means 75%. So correct option is A

Similarly for 2nd question 1-1/9=8/9 or 89% at least. Range is 63.6+3*2.5 or 71.1. option A

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