Question

1) The scores on a test administered to prospective employees are normally distributed with amean of 100 and a standard deviation of 15 (use this infornation for 1-6). About what percent of the scores are between 70 and 130?
2) About what percent of the scores are between 85 and 130?
3) About what percent of the scores are over 115?
4) About what percent of the scores are lower than 85 or higher than 115?
5) If 80 people take the test, how many would you expect to score higher than 130?

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6) If 75 people take the test, how many would you expect to score lower than 85?

Answer 1

This is a normal distribution question with

a) P(70.0 < x < 130.0)=?

This implies that

P(70.0 < x < 130.0) = P(-2.0 < z < 2.0) = P(Z < 2.0) - P(Z < -2.0)

P(70.0 < x < 130.0) = 0.9772498680518208 - 0.022750131948179195

P(70.0 < x < 130.0) = 0.9545

b) P(x > 115.0)=?

The z-score at x = 115.0 is,

z = 1.0

This implies that

P(x > 115.0) = P(z > 1.0) = 1 - 0.8413447460685429

P(x > 115.0) = 0.1587

c)
P(85.0 < x < 130.0)=?

This implies that

P(85.0 < x < 130.0) = P(-1.0 < z < 2.0) = P(Z < 2.0) - P(Z < -1.0)

P(85.0 < x < 130.0) = 0.9772498680518208 - 0.15865525393145707

P(85.0 < x < 130.0) = 0.8186

d)

P(X < 85.0 or X > 115.0)=?

This implies that

P(X < 85.0 or X > 115.0) = P(z < -1.0 or z > 1.0) = 0.3173

e)

Sample size (n) = 80

Since we know that

P(x > 130.0)=?

The z-score at x = 130.0 is,

z = 17.888

This implies that

P(x > 130.0) = P(z > 17.888) = 1 - 1.0

P(x > 130.0) = 0

f)

Sample size (n) = 75

Since we know that

P(x < 85.0)=?

The z-score at x = 85.0 is,

z = -8.66

This implies that

P(x < 85.0) = P(z < -8.66) = 0

PS: you have to refer z score table to find the final probabilities.

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