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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?

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frika

Answer:

0.0274

Step-by-step explanation:

The mean is [tex]\mu =114.8[/tex] and the standard deviation is [tex]\sigma =13.1.[/tex]

Calculate

[tex]Z=\dfrac{X-\mu}{\sigma}[/tex]

for [tex]X=140:[/tex]

[tex]Z=\dfrac{140-114.8}{13.1}\approx 1.9237.[/tex]

If [tex]X\sim N(114.8,\ 13.1),[/tex] then [tex]Z\sim N(0,1)[/tex]

and

[tex]Pr(X>140)=Pr(Z>1.9237).[/tex]

Use table for normal distribution probabilities to get that

[tex]Pr(Z>1.9237)=1-Pr(Z\le 1.9237)=1-0.9726=0.0274.[/tex]

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

=[tex]P(X>140)[/tex]

=[tex]P(Z>\frac{140-114.8}{13.1} =P(Z>1.923)\\[/tex]

=0.5-0.4726

=0.0274

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