Question

xa3 led that their annual incomes from employment n industry during the day were normally distributed a standard deviation of $3,000. 14).A sample of 500 evening students revea with a mean income of $30,000 and 5-000 (i) 250 students earned more than $30,000 (ii) 314 students earned between $27,000 and $33,000. (iii) 239 students earned between $24,000 and $30,000. A) and(i) are correct statements but not (ii). (i) and (iii) are correct statements but not (ii). ) and (iii) are correct statements but not (i). D) (i), (ii), and (iii) are all correct statements. E) (i) is a correct statement but not (ii) or (ii). 15) 15) 1 wo business major students, in two different sections of economics, were comparing test scores. The following gives the students' scores, class mean, and standard deviation for each section. Section Score 84 75 75 2 60 (i) The student from Section 2 scored better compared to the rest of their section. (ii) The z-score of the student from section 1 is 1.28. (ii) The z-score of the student from section 2 is 1.87. A) (i) and (ii) are correct statements but not (ii). B)(ii) and (iii) are correct statements but not (i). C) fi), (ii), and (iii) are all correct statements D) (i) is a correct statement but not (ii) or (iii). B) (G) and (ii) are correct statements but not (ii) →( . 16) For a standard normal distribution, what is the probail ity that z is greater than 1.75 16) A) 0.0401 B) 0.4599 0.9599 D) 0.0459 17) A sample of 500 evening students revealed that their annual incomes were normally 17) distributed with a mean income of $50,000 and a standard deviation of $4,000. How many students earned less than 845,000? A) 53 C) 35 D) 197 18) What is the proportion of the total area under the normal curve within plus and minus 18) two standard deviations of the mean? A) 95% B) 34% 8% D) 99.7%

Answer 1

14)i) P(X > 30000)

= P((X - $\mu$ )/ $\sigma$ > (30000 - $\mu$ )/ $\sigma$ )

= P(Z > (30000 - 30000)/3000)

= P(Z > 0)

= 1 - P(Z < 0)

= 1 - 0.5 = 0.5

Expected no of students = 500 * 0.5 = 250

ii) P(27000 < X < 33000)

= P((27000 - 30000)/3000 < Z < (33000 - 30000)/3000)

= P(-1 < Z < 1)

= P(Z < 1) - P(Z < -1)

= 0.8413 - 0.1587

= 0.6826

Expected no of students = 500 * 0.6826 = 341.3

iii) P(24000 < X < 30000)

= P((24000 - 30000)/3000 < Z < (30000 - 30000)/3000)

= P(-2 < Z < 0)

= P(Z < 0) - P(Z < -2)

= 0.5 - 0.0228

= 0.4772

Expected no of students = 500 * 0.4772 = 239

Option - B) (i) and (iii) are correct statements but not (ii).

15) For section - 1

z-score = (x - $\mu$ )/ $\sigma$

             = (84 - 75)/7 = 1.2857

For section - 2

z-score = (x - $\mu$ )/ $\sigma$

            = (75 - 60)/8 = 1.875

Option - C) (i), (ii), and (iii) all are correct statements.

16) P(Z > 1.75)

     = 1 - P(Z < 1.75)

     = 1 - 0.9599

     = 0.0401

Option - A is correct

17) P(X < 45000)

= P((X - $\mu$ )/ $\sigma$ < (45000 - $\mu$ )/ $\sigma$ )

= P(Z < (45000 - 50000)/4000)

= P(Z < -1.25)

= 0.1056

Expected no of students = 500 * 0.1056 = 53

18) P(-2 < Z < 2)

= P(Z < 2) - P(Z < -2)

= 0.9772 - 0.0228

= 0.9544 = 95.44%

Option - A is correct