Mathematics, 24.02.2020 19:47 wi8wuwj283jendjdudjd

Based on the Normal model N(100,17) describing IQ scores, what percent of individuals would be expected to have scores a) over 80? b) under 89? c) between 111 and 131? d) over 128?

Answers: 3

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Answer from: pmdare2551

a) 88.10% of students would be expected to score over 80

b) 25.78% of students would be expected to score under 89.

c) 22.34% of students would be expected to score between 111 and 131.

d) 4.95% of students would be expected to score over 128.

Step-by-step explanation:

Problems of normally distributed samples are solved using 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 problem, we have that:

$\mu = 100, \sigma = 17$

a) over 80?

1 subtracted by the pvalue of Z when X = 80.

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

$Z = \frac{80 - 100}{17}$

Z = -1.18

Z = -1.18 has a pvalue of 0.1190

1 - 0.1190 = 0.8810

88.10% of students would be expected to score over 80

b) under 89?

Pvalue of Z when X = 89. So

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

$Z = \frac{89 - 100}{17}$

Z = -0.65

Z = -0.65 has a pvalue of 0.2578.

25.78% of students would be expected to score under 89.

c) between 111 and 131?

Pvalue of Z when X = 131 subtracted by the pvalue of Z when X = 111.

X = 131

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

$Z = \frac{131 - 100}{17}$

Z = 1.82

Z = 1.82 has a pvalue of 0.9656.

X = 111

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

$Z = \frac{111 - 100}{17}$

Z = 0.65

Z = 0.65 has a pvalue of 0.7422.

0.9656 - 0.7422 = 0.2234

22.34% of students would be expected to score between 111 and 131.

d) over 128?

1 subtracted by the pvalue of Z when X = 128. So

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

$Z = \frac{128 - 100}{17}$

Z = 1.65

Z = 1.65 has a pvalue of 0.9505

1 - 0.9505 = 0.0495

4.95% of students would be expected to score over 128.

Answer from: Quest

283 square feet

step-by-step explanation:

good luck

Due today (05.06)use a net to find the surface area of the right triangular prism shown below: 88

Answer from: Quest

1. x> -5

2. x< 8

step-by-step explanation:

1. 2x> -10

divide by 2 from both sides of equation.

2x/2> -10/2

simplify, to find the answer.

-10/2=-5

x> -5 is the correct answer.

2. 3x< 24

divide by 3 from both sides of equation.

3x/3< 24/3

simplify, to find the answer.

24/3=8

x< 8 is the correct answer.

i hope this you, and have a wonderful day!

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Based on the Normal model N(100,17) describing IQ scores, what percent of individuals would be expec...