Question

11. Given that the ages (in years) of the reside

nts in a city are normally

distributed with a population mean of 37 and a population standard deviation of 13. The probability that the age of a randomly selected person

from this city is between 30 and 44 is closest to

(a) 0.21

(b) 0.54

(c) 0.41

(d) 1.0

(e)None of these.

12. Weights of newborn babies in a certain state have normal distribution with

mean 5.33 lb and standard deviation 0.65 lb. A newborn weighing less than

4.85 lb is considered to be

at risk, that is, has a higher mortality rate.

(a) A baby just born in this state is picked at random. The probability that

the baby is at risk is about

(a) 0.43

(b) 0.33

(c) 0.23

d) 0.13

(e) 0.53

(b) The hospital wants to take pictures of the heaviest 10% of the newborn

babies. The minimum weight (in lbs) required for a picture to be taken is

about

(a) 5.89

(c) 6.16

(c) 9.12

(d) 8.12

(e) 7.27

Answer 1

Solution( 11 ) :

We are given that : the ages (in years) of the residents in a city are normally distributed with a population mean of 37 and a population standard deviation of 13.

That is : Mean = and Standard Deviation =

We have to find the probability that the age of a randomly selected person from this city is between 30 and 44 .

That is : P( 30 < X < 44 ) =...........?

Thus find z scores for x = 30 and for x = 44

Thus we get :

P( 30 < X < 44 ) =P( -0.54 < Z < 0.54 )

P( 30 < X < 44 ) =P( Z < 0.54 ) - P( Z < -0.54 )

Look in z table for z = -0.5 and 0.04 as well as for z = 0.5 and 0.04

and find corresponding area.

P( Z < -0.54 ) = 0.2946

P( Z < 0.54) = 0.7054

Thus

P( 30 < X < 44 ) =P( Z < 0.54 ) - P( Z < -0.54 )

P( 30 < X < 44 ) = 0.7054   - 0.2946

P( 30 < X < 44 ) = 0.4108

P( 30 < X < 44 ) =0.41

Thus correct option is : c) 0.41 .