Asked by eric
For males in a certain town, the systolic blood pressure is normally distributed with a mean of 125 and a standard deviation of 9. If 434 males from the town are randomly selected to participate in a study, how many of them would be expected to have a systolic blood pressure between 106 and 136, to the nearest whole number?
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We want P(106 ≤ X ≤ 136) for X ~ N(125, 9^2).
Compute z-scores:
z1 = (106 − 125)/9 = −19/9 ≈ −2.111
z2 = (136 − 125)/9 = 11/9 ≈ 1.222
Proportion = Φ(1.222) − Φ(−2.111) ≈ 0.889 − 0.0174 ≈ 0.8716.
Expected count = 0.8716 × 434 ≈ 378 (nearest whole number).
Answer: 378.
Compute z-scores:
z1 = (106 − 125)/9 = −19/9 ≈ −2.111
z2 = (136 − 125)/9 = 11/9 ≈ 1.222
Proportion = Φ(1.222) − Φ(−2.111) ≈ 0.889 − 0.0174 ≈ 0.8716.
Expected count = 0.8716 × 434 ≈ 378 (nearest whole number).
Answer: 378.
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