Question

Use this information to answer Question 1 to Question 9: Z is a Standard Normal variable. 1. (1 point) P(Z ≤ 1.2) = (a) 0.8849 (b) 0.1151 (c) 0.6482 (d) 0.3685 2. (1 point) P(Z ≥ 1) = (a) 0.8413 (b) 0.6821 (c) 0.3586 (d) 0.1587 3. (1 point) P(−0.4 < Z < 0.5) = (a) 0.8962 (b) 0.3469 (c) 0.8762 (d) 0.2562 Page 1 of 9 BUSI 1013 Assignment 003 Spring 2020 4. (1 point) P(Z < −0.9) = (a) 0.4524 (b) 0.8721 (c) 0.1841 (d) 0.4691 5. (1 point) P(Z > −0.8) = (a) 0.3658 (b) 0.7881 (c) 0.3527 (d) 0.7372 6. (1 point) Find z such that P(Z < z) = 0.4. (a) -0.25 (b) -0.75 (c) 0.25 (d) 0.75 7. (1 point) Find z such that P(Z > z) = 0.4. (a) 0.75 (b) -0.25 (c) -0.75 (d) 0.25 8. (1 point) Find z such that P(Z > z) = 0.6 (a) 0.75 (b) 0.25 (c) -0.25 (d) -0.75 9. (1 point) Find z such that P (−z < Z < z) = 0.6. (a) -1.16 (b) -0.84 (c) 1.16 (d) 0.84 Page 2 of 9 BUSI 1013 Assignment 003 Spring 2020 Use this information to answer Question 10 to Question 13: Let X be the height of adults in Toronto. Height of Adults in Toronto is normally distributed with mean 170 cms and standard deviation 4 cms. 10. (1 point) What is the probability that height of adults in Toronto is less than 174 cms? (a) 0.8410 (b) 0.1590 (c) 0.4528 (d) 0.1764 11. (1 point) What is the probability that height of adults in Toronto is more than 168 cms? (a) 0.244 (b) 0.127 (c) 0.633 (d) 0.691 12. (1 point) What is the probability that height of adults in Toronto is less than 168 cms? (a) 0.376 (b) 0.556 (c) 0.309 (d) 0.836 13. (1 point) What is the probability that height of adults in Toronto is between 165 cms and 175 cms? (a) 0.266 (b) 0.788 (c) 0.166 (d) 0.676 Use this information to answer Question 14 to 16: The cost of monthly commute in Toronto is normally distributed with mean $90 and standard deviation of $10. 14. (1 point) What is the probability that the cost of monthly commute of a person is more than $87? (a) 0.618 (b) 0.376 (c) 0.973 (d) 0.255 Page 3 of 9 BUSI 1013 Assignment 003 Spring 2020 15. (1 point) What is the probability that the cost of monthly commute is less than $103? (a) 0.562 (b) 0.265 (c) 0.903 (d) 0.169 16. (1 point) What is the probability that the cost of monthly commute of a person is between $78 and $95? (a) 0.756 (b) 0.576 (c) 0.265 (d) 0.832 Use this information to answer Question 17 to Question 18: Yorkville University students scored an average of 145 and standard deviation of 3 on IQ Test. University of Toronto students scored an average of 150 and standard deviation of 9 on IQ test. Assume that the scores of Yorkville University students and University of Toronto students are normally distributed. 17. (1 point) What percent of Yorkville University students scored less than 140? (a) 95.2% (b) 4.8% (c) 75.2% (d) 34.6% 18. (1 point) What percent of University of Toronto students scored less than 140? (a) 26.6% (b) 37.5% (c) 13.3% (d) 86.2% 19. (2 points) Variable X is a normal random variable with standard deviation 3. If the probability that X is less than 16 is 0.84, then what is the mean of X ? (a) 13 (b) 12 (c) 16 (d) 14 Page 4 of 9 BUSI 1013 Assignment 003 Spring 2020 20. (2 points) Variable X is a normal random variable with mean 100. If the probability that X is greater than 90 is 0.84, then what is the standard deviation of X ? (a) 16 (b) 15 (c) 13 (d) 10 Part B: Sampling Distribution Use this information to answer Question 21 to Question 23: Sample of 100 customers is selected from an account receivable portfolio. The population mean account balance is $1500. The population standard deviation is known to be $90. 21. (3 points) What is the probability that mean account balance of the sample is more than $1510? (a) 0.456 (b) 0.867 (c) 0.133 (d) 0.346 22. (3 points) What is the probability that the mean account balance of the sample is less than $1495? (a) 0.289 (b) 0.256 (c) 0.572 (d) 0.452 23. (3 points) What is the probability that the mean account balance of the sample is between $1490 and $1505? (a) 0.672 (b) 0.976 (c) 0.814 (d) 0.578 Use this information to answer Question 24 to Question 26: Sample of 100 customers is selected from an account receivable portfolio and the sample mean account balance is $1500. The population standard deviation is known to be $90. Page 5 of 9 BUSI 1013 Assignment 003 Spring 2020 24. (3 points) Find the 95% confidence interval for the mean account balance of the population? (a) (1476.3,1490.7) (b) (1482.4,1517.6) (c) (1476.4,1524.7) (d) (1498.4,1534.8) 25. (2 points) What is the margin of error associated with 95% confidence interval? (a) 12.6 (b) 14.2 (c) 18.4 (d) 17.6 26. (3 points) Suppose the sample was of size 150 instead of 100. Rest of the values remain the same. What would be the 95% confidence interval for the mean account balance of the population? (a) (1485.6,1514.4) (b) (1456.7,1567.4) (c) (1434.7,1565.3) (d) (1450.4,1549.6) Use this information to answer Question 27 to Question 28: A sample of 75 calls is monitored at an in-bound call centre and the average length of the calls is 4 minutes. The population standard deviation is unknown. The sample standard deviation is found to be 4 minutes. 27. (3 points) What is the 95% confidence interval for the average length of inbound calls? (a) (3.56,4.44) (b) (3.45,4.55) (c) (3.25,4.75) (d) (3.08,4.9) 28. (2 points) What is the margin of error associated with 95% confidence interval? (a) 0.67 (b) 0.92 (c) 0.45 (d) 1.34

Answer 1

Q1:

= P(z < 1.2)

Using excel function:

= NORM.S.DIST(1.2, 1)

= 0.8849

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Q2:

= P(z > 1)

= 1 - P(z < 1)

Using excel function:

= 1 - NORM.S.DIST(1, 1)

= 0.1587

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Q3:

= P(-0.4 < z < 0.5)

= P(z < 0.5) - P(z < -0.4)

Using excel function:

= NORM.S.DIST(0.5, 1) - NORM.S.DIST(-0.4, 1)

= 0.3469

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Q4:

= P(z < -0.9)

Using excel function:

= NORM.S.DIST(-0.9, 1)

= 0.1841

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Q5:

= P(z > -0.8)

= 1 - P(z < -0.8)

Using excel function:

= 1 - NORM.S.DIST(-0.8, 1)

= 0.7881

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Q6:

P(Z < z) = 0.4

Z score at p = 0.4 using excel = NORM.S.INV(0.4) = -0.25

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Q7:

P(Z > z) = 0.4

= 1 - P(Z < z) = 0.4

= P(Z < z) = 0.6

Z score at p = 0.6 using excel = NORM.S.INV(0.6) = 0.25

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Q8:

P(Z > z) = 0.6

= 1 - P(Z < z) = 0.6

= P(Z < z) = 0.4

Z score at p = 0.4 using excel = NORM.S.INV(0.4) = -0.25

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Q9:

Proportion in the middle = 0.6

Proportion on left and right side of normal curve = 0.2

Z score at p = 0.2 using excel = NORM.S.INV(0.2) = 0.84

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