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Natali5045456 [20]
1 year ago
8

Question regarding a Normal Distribution chart: Ten thousand adults were given a literacy test. The results were nearly normally

distributed (μ = 75, σ​ = 15).
A. About how many scored between 60 and 80?


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


C. Find the percentile rank of the person who scored 88.
Mathematics
1 answer:
nirvana33 [79]1 year ago
4 0

Answer:

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 \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

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:

\mu = 75, \sigma = 15

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

Z = \frac{X - \mu}{\sigma}

Z = \frac{80 - 75}{15}

Z = 0.33

Z = 0.33 has a pvalue of 0.6293

X = 60

Z = \frac{X - \mu}{\sigma}

Z = \frac{60 - 75}{15}

Z = -1

Z = -1 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.

Z = \frac{X - \mu}{\sigma}

1.037 = \frac{X - 75}{15}

X - 75 = 1.037*15

X = 90.6

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

Z = \frac{X - \mu}{\sigma}

Z = \frac{88 - 75}{15}

Z = 0.87

Z = 0.87 has a pvalue of 0.8078

So, rounding up, the 81th percentile.

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