Question

Suppose that diastolic blood pressure in hypertensive women is normally distributed with a mean of 100 mg Hg and a standard deviation of 14mm Hg. If 10 hypertensive women were randomly selected from this population, what is the probability that exactly 5 will have diastolic blood pressure less than 95 mgHg?

Answer 1

P[Z < -3.572]

Explanation:

Given that;

`\mu = 100\\ \sigma=14`

A sample of size (n) = 10

Let
`\bar x` be mean of the sample

So the sampling distribution of
`\bar x` = Mean
`\mu_ {\bar x} = \mu =100`

SD
`\sigma _( \bar x) =(\sigma)/(√(n))`

`\sigma _( \bar x) =(14)/(√(10))`

=4.4

To calculate:
`P( \bar x < 95)`; we have:

`= P[((\bar x - \mu_(\bar x)))/((\sigma _(\bar x)<(95-100))/(14) ) ]`

= P [Z < -3.572]