Question

Assume that the ages for first marriages are normally distributed with a mean of 26 years and a standard deviation of 4 years. What is the probability that a person getting married for the first time is in his or her twenties? (Round your answer to 4 decimal places.)

Answer 1

Answer:

0.7743

Step-by-step explanation:

Mean of age = u = 26 years

Standard Deviation = $\sigma$ = 4 years

We need to find the probability that the person getting married is in his or her twenties. This means the age of the person should be between 20 and 30. So, we are to find P( 20 < x < 30), where represents the distribution of age.

Since the data is normally distributed we can use the z distribution to solve this problem. The formula to calculate the z score is:

$z=\frac{x-u}{\sigma}$

20 converted to z score will be:

$z=\frac{20-26}{4}=-1.5$

30 converted to z score will be:

$z=\frac{30-26}{4}=1$

So, now we have to find the probability that the z value lies between -1.5 and 1.

P( 20 < x < 30) = P( -1.5 < z < 1)

P( -1.5 < z < 1 ) = P(z < 1) - P(z<-1.5)

From the z-table:

P(z < 1) = 0.8413

P(z < -1.5) =0.067

So,

P( -1.5 < z < 1 ) = 0.8413 - 0.067 = 0.7743

Thus,

P( 20 < x < 30) = 0.7743

So, we can conclude that the probability that a person getting married for the first time is in his or her twenties is 0.7743