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labwork [276]
3 years ago
13

A woman is randomly selected from the 18-24 age group. For women of this group, systolic blood pressures (in mm Hg) are normally

distributed with a mean of 114.8 and a standard deviation of 13.1. What is the probability this woman has a systolic blood pressure greater than 140?
Mathematics
2 answers:
jonny [76]3 years ago
8 0

Answer:

0.0224

Step-by-step explanation:

Given that systolic blood pressures of women are normally distributed.

IF X is the systolic bp of women X is N(114.8, 13.1)

Required probability

=  the probability this woman has a systolic blood pressure greater than 140

=P(X>140)

=P(Z>\frac{140-114.8}{13.1} =P(Z>1.923)\\

=0.5-0.4726

=0.0274

dybincka [34]3 years ago
7 0

Answer:

0.0274

Step-by-step explanation:

The mean is \mu =114.8 and the standard deviation is \sigma =13.1.

Calculate

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

for X=140:

Z=\dfrac{140-114.8}{13.1}\approx 1.9237.

If X\sim N(114.8,\ 13.1), then Z\sim N(0,1)

and

Pr(X>140)=Pr(Z>1.9237).

Use table for normal distribution probabilities to get that

Pr(Z>1.9237)=1-Pr(Z\le 1.9237)=1-0.9726=0.0274.

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