ten thousand adults were given a literacy test. the results were nearly normally distributed (µ = 75, δ = 15). (20 points) a. about how many scored between 60 and 80?

a) 4,706 scored between 60 and 80

b) A score of 90.6.

c) 81th percentile.

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean  and standard deviation , the score 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 question, we have that:

A. About how many scored between 60 and 80?

Proportion:

This is the pvalue of Z when X = 80 subtracted by the pvalue of Z when X = 60. So

X = 80

has a pvalue of 0.6293

X = 60

has a pvalue of 0.1587

0.6293 - 0.1587 = 0.4706

How many?

0.4706 out of 10,000, so

0.4706*10000 = 4706

4,706 scored between 60 and 80.

B. Approximately what score did only the top 15% exceed?

The 85th percentile, which is X when Z has a pvalue of 0.85. So X when Z = 1.037.

A score of 90.6.

C. Find the percentile rank of the person who scored 88.

This is the pvalue of Z when X = 88. So

has a pvalue of 0.8078

So, rounding up, the 81th percentile.

__if the first digit is 1, then the second digit is 4 times that: 4.so 14 and 41. in both cases, one digit is 4 times the other.if the first digit is 2, then the second digit is 4 times that: 8so 28 and 82if the first digit is 3, then 3 * 4 = 12, which would not be a valid digit. in fact, 1 and 2 are all the digits that you could multiply by 4 to still get a valid number.14 + 41 + 28 + 82 = 165