Heights of fourth graders are Normally distributed with a mean of 52 inches and a standard deviation of 3.5 inches. What percentage of fourth graders are between 50 inches and 56 inches? A. 68.13% B. 27.42% C. 59.01% D. 95.27% E. 86.43%

C. 59.01%

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

What percentage of fourth graders are between 50 inches and 56 inches?

This is the pvalue of Z when X = 56 subtracted by the pvalue of Z when X = 50.

X = 56

has a pvalue of 0.8729

X = 50

has a pvalue of 0.2843.

0.8729 - 0.2843 = 0.5886, which is close to 59.01%.

So the correct answer is:

C. 59.01%

answer is 67%

step-by-step explanation:

step-by-step explanation:

12: 00 p.m. to 7: 00 p.m. is a total of 7 hours.

4: 00 p.m. to 7: 00 p.m. is a total of 3 hours.

this means that from 4: 00 p.m to 7: 00 p.m. this represents 3/7 of what was lost from 12: 00 p.m. to 7: 00 p.m.

after knowing this, after doing 11,600 - 6,800 = 4,800

4,800 represents 3/7 of what was lost for 7 hours.

we then divide 4,800 by 3 which gets 1,600.

because we know that 1,600 represents 1/7 of how much water was lost, we just have to multiply by 7, which gets 11,200.   after adding 11,200 to how much water was left at 7: 00 p.m. which was 6,800. 11,200 + 6,800 = 18,000

this means that the pool originally had 18,000 gallons in the pool.

you can also check your work by doing 18,000 - 4(1600), which becomes           1800 - 6400 which gets you 11,600.