Question 1 10 points Heights of adult women have a mean of 63.6 in. and a standa

Question

Question 1 10 points Heights of adult women have a mean of 63.6 in. and a standard deviation of 2.5 in. a. What does Chebyshev's Theorem say about the percentage of women with heights between 58.6 in. and 68.6 in? 2.5 pts. At least 75% of woman's height is between 58.6 and 68.6in. a. what interval represents 80% of the data? 2.5pts 80% of values is within 58.010 and 69.190 Elephants have the longest pregnancy of all mammals. One species of elephant has a mean gestation period of 520 days and standard deviation of 31 days. Pregnancy length follows an approximately normally distribution. Use the 68%-95%-99.7% Rule to approximate each of the following. 1.25 pts. Each The longest 16% of all elephant pregnancies last at least how many days? 551 days The middle 68% of all elephant pregnancies last between 489 and 551 days and. days. days. Only 2.5% of a elephant pregnancies last longer than What percent of elephant pregnancies last less than 458 days?.. 2.28%

Explanation / Answer

Answer to the questions as below:

a.

So, 58.6 and 68.6 are 2 deviations above and below mean.
Chebeshev' theorm states that 1-1/K^2 is the area under cover with k deviations from the mean.
So, the %age of women with heights between 58.6 and 68.6 is 1 - 1/2^2 = 1-1/4 = 75%

b. 80% is 1-1/(5)

So, k = sqrt(5) = 2.236
Hence, the 63.6 +/- 2.236*2.5 = 58.010 to 69.19

c.

Mean = 520
Stdev = 31
So, 1,2,3 deviataions way from mean cover 68%-95%-99.7% of area
The longest 16% of all elephant preg. last at least?
68% = 34%*2 . So, 34 % each side of mean
Hence, to get cumulative area of 84% , .5+.34 move 1 deviation above mean.
Hen,ce 520+31 should be the score. 551 datys

Middle 68% i s 1 deviations away from mean
520+/- 31 or 489 to 551 days

2.5% area can be 2 deviation above mean.

So, 520+31*2 = 582 days

So, P(X<458) = P(Z< 458-551 / 31) = P(Z<3) = .02275 or .0228 or 2.28%

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