Question

Women have a pulse rates that are normally distributed with a mean of 77.5 beats per minute and a standard deviation of 11.6 per minute.

a) Dr. Purets sees exactly 25 female patients each day. Find the probability that 25 randomly selected have a mean pulse rate between 70 beats per minute and 85 beats per minute.

b) Two percent of the women with the highest pulse rates have a rate greater than what number?

Answer 1

Solution :

Given that,

mean = = 77.5

standard deviation = = 11.6

n = 25

= = 77.5

= / n = 11.6 / 25 = 2.32

1)

P(70 < < 85) = P((70 - 77.5) / 2.32<( - ) / < (85 - 77.5) / 2.32))

= P(-3.23 < Z < 3.23)

= P(Z < 3.23) - P(Z < -3.23) Using standard normal table,  

= 0.9994 - 0.0060

= 0.9934

Probability = 0.9934

2)

P( Z > z ) = 0.02 %

1 - P( Z < z ) = 0.02

P( Z <  ) = 1 - 0.02

P( Z < z ) = 0.98

P( Z < 2.054 ) =

z = 2.054

Using z - score formula,

X = z * +

= 2.054 * 2.32 + 77.5  

= 82.3