Question

It was noted that women have normally distributed pulse rates with a mean of 75.0 bpm and a standard deviation of 10 bpm

• A) Find the probability that 1 randomly selected woman has a pulse rate greater than 80 bpm
• (B) Find the value of the pulse rate ranked as the top 1%
• (C) Find the value of the pulse rate ranked as the bottom 2%
• (D) Find the value of the pulse rate ranked between top 5% and bottom 5%

Answer 1

Here we have : Mean = = 75, standard deviation = 10

A) Here we have to find the probability that 1 randomly selected woman has a pulse rate greater than 80 bpm.

p ( x > 80 )

= p ( z > 0.5 )
= 1 - p ( z 0.5 )

= 1 - 0.6915 -------------( using excel formula "=norm.s.dist( 0.5, 1) " )

= 0.3085

b) Here we need to find the value of the pulse rate ranked as the top 1%.

We express thin in the form of z probability equation.

p ( Z > z ) = 0.01

1 - p ( Z z ) = 0.01

p ( Z z ) = 0.99

z = 2.33 -----( using excel formula " =norm.s.inv( 0.99 ) " )

x = 98.3

c) Here we need to find, the value of the pulse rate ranked as the bottom 2% .

We express thin in the form of z probability equation.

p ( Z < z ) = 0.02

z = -2.05 -----( using excel formula " =norm.s.inv( 0.02 ) " )

x = 54.5

d)

Here we need to find the value of the pulse rate ranked between top 5% and bottom 5%.

Pulse rate value for top 5% is given by,

p ( Z > z ) = 0.05

1 - p ( Z z ) = 0.05

p ( Z z ) = 0.95

z = 1.645 -----( using excel formula " =norm.s.inv( 0.95 ) " )

x = 91.45

Pulse rate value for bottom 5% is given by,

p ( Z < z ) = 0.05

z = -1.645 -----( using excel formula " =norm.s.inv( 0.05 ) " )

x = 58.55

Hence pulse ranked between two values 58.55 and 91.45 .