The mean number of words per minute (WPM) read by sixth graders is 89 with a standard deviation of 16 WPM. If 66 sixth graders are randomly selected, what is the probability that the sample mean would differ from the population mean by more than 5.09 WPM?

0.98% probability that the sample mean would differ from the population mean by more than 5.09 WPM.

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

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.

Central limit theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean and standard deviation , a large sample size can be approximated to a normal distribution with mean

In this problem, we have that:

What is the probability that the sample mean would differ from the population mean by more than 5.09 WPM?

Either it differs less than 5.09 or equal, or it differs by more than 5.09. The sum of these probabilities is 100%.

So we first find the probability that the sample mean would differ from the population mean by less than 5.09 WPM, and then we subtract 100 by this probability.

Differ less than 5.09

pvalue of Z when X = 89 + 5.09 = 94.09 subtracted by the pvalue of Z when X = 89 - 5.09 = 83.91

X = 94.09

By the Central Limit Theorem

has a pvalue of 0.9951.

X = 83.91

has a pvalue of 0.0049.

0.9951 - 0.0049 = 0.9902 = 99.02% probability that it differs less than 5.09

Differ more than 5.09

100 - 99.02 = 0.98% probability that the sample mean would differ from the population mean by more than 5.09 WPM.

something that characterises a right triangle is the fact that we may use pythagoras' theorem to find any one of its sides given that we have the length of the other two sides. here, we have the length of the hypotenuse, √(87) yd, and one other side, √(23) yd, thus we can easily find the length of the remaining side using pythagoras' theorem as defined below:

c^2 = a^2 + b^2, where c is the length of the hypotenuse, and a and b are the lengths of the other two sides.

thus, given that c = √(87) and a = √(23), we may substitute these into the formula we defined above to get:

c^2 = a^2 + b^2

(√87)^2 = (√23)^2 + b^2 (substitute c = √87 and a = √23 into the equation)

87 = 23 + b^2 (evaluate (√87)^2 = 87 and (√23)^2 = 23 - remember that the square root of a number, squared will equal that very same number)

64 = b^2 (subtract 23 from both sides)

√64 = b (take the square root of both sides)

8 = b (evaluate √64 = 8)

therefor, the length of the unknown leg in the right triangle is 8 yd (first answer).