Question

Questions 15 –17

Personal incomes in a city have an average of $80,000 and standard deviation of $20,000.

15. The percent of personal incomes over $100,000 is

(a) 0.05 (b) 0.1 (c) 0.1587 (d) 0.1841 (e) Not enough information to calculate.

16. In a random sample of 100 people, the probability that the average income in the sample is over $82,000 is

(a) 0.05 (b) 0.1   (c) 0.1587 (d) 0.1841 (e) Not enough information to calculate.

17. In another random sample of 400 people, the probability that the average income in this sample is at least $3,000 more than the sample average income in Question 16 is

(a) 0.03 (b) 0.05 (c) 0.07 (d) 0.09 (e) Not enough information to calculate.

Answer 1

Solution:

Given, the Normal distribution with,

   = 80000

= 20000

15)  P(X > 100000) = P[(X - )/ >  (100000 - 80000)/ 20000]

= P[Z > 1.00]

= P[Z < -1.00]

= 0.1587 ..( use z table)

Answer :   0.1587

16)n = 100

Let be the mean of sample.

The sampling distribution of the is approximately normal with

Mean() = = 80000

SD() =    = 20000/​100 = 2000

Find P( > 82000)

= P[( - )/ >  (82000 - )/]

= P[Z > (82000 - 80000)/2000 ]

= P[Z > 1.00]

  = P[Z < -1.00]

= 0.1587 ..( use z table)

Answer :   0.1587

17) n = 400

Find P( average income in this sample is at least $3,000 more than the sample average income in Question 16)

But , in question 16 , the sample average income is not given .

So ,

Answer :  (e) Not enough information to calculate.