Question

The blood pressure of the 1000 males in the study described in #4 had their blood pressures recorded. The systolic blood pressures of the males have an approximately normal distribution with a mean LaTeX: \mu=125\:millimetersμ = 125 m i l l i m e t e r s and standard deviation LaTeX: \sigma=13\:millimetersσ = 13 m i l l i m e t e r s.

A: Estimate the number of men in the study whose blood pressure was between 99 and 151 millimeters.

B: Estimate the number of men in the study whose blood pressure was greater than 99 millimeters.

C: Estimate the number of men in the study whose blood pressure was between 99 and 138 millimeters

Answer 1

Given = 125 mm, = 13 mm

To find the probability, we need to find the z scores.

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(A) Between 99 and 151 = P(99 < X < 151) = P(151) - P(99)

For P(151) ; z = (151 - 125) / 13 = 2. The p value at this score is = 0.9772

For P(200) ; z = (99 - 125) / 13 = -2.0. The p value at this score is = 0.0228

Therefore the required probability is 0.9772 – 0.0228 = 0.9544

Therefore the number of men between these two values = 1000 * 0.9544 = 954.4        955

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(B) For P (X > 99) = 1 - P (X < 99), as the normal tables give us the left tailed probability only.

From (A) The probability for P(X < 99) from the normal distribution tables is = 0.0228

Therefore the probability for P(X > 99) = 1 - 0.0228 = 0.9772

Therefore the number of men between these two values = 1000 * 0.9772 = 977.2       977

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(C) Between 99 and 138 = P(99 < X < 138) = P(138) - P(99)

For P(138) ; z = (138 - 125) / 13 = 1. The p value at this score is = 0.8413

For P(99) ; z = (99 - 125) / 13 = -2.0. The p value at this score is = 0.0228

Therefore the required probability is 0.8413 – 0.0228 = 0.8185

Therefore the number of men between these two values = 1000 * 0.8185 = 818.5    819

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Please consider the final answers which are integers, only if required.