STAT 245 Lecture Notes - Lecture 5: Random Variable, Simple Random Sample, Old Age Security
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STATISTICS 244.3
Exercises #5 - SOLUTIONS
PROBLEM A: Suppose Z is a Standard Normal random variable. That is, Z ~ N (0, 1). For
each question, draw an appropriate picture and then find the numerical answer.
[1]. Find each of the following probabilities.
(a). P[Z ≤ 2.07] = 0.9808
(b). P[Z > -0.69] = 1 - P[Z ≤ -0.69]
= 1 – 0.2451 = 0.7549
(c). P[-1.26 <Z ≤+1.71]
= P[Z ≤+1.71]-P[Z <-1.26]
= 0.9564 – 0.1038= 0.8526
[2]. Find the value z in each case.
(a). P[Z ≤ z] = 0.7088 => z = +0.55
(b). P[Z > z] = 0.8708 => P[Z ≤ z] = 0.1292
=> z = - 1.13
(c). P[-z ≤ Z ≤ +z] = 0.6630
=> P[Z ≤ +z] = 0.3315 + 0.5000 = 0.8315
=> z = +0.96
or P[-z ≤ Z ≤ +z] = 0.6630
=> P[Z ≤ -z] = ½(1-0.6630) = 0.1685
=> P[Z ≤ +z] = 0.6630 + 0.1685 = 0.8315
=> z = +0.96
Document Summary
Problem a: suppose z is a standard normal random variable. For each question, draw an appropriate picture and then find the numerical answer. Find the value z in each case. (a). P[z z] = 0. 7088 => z = +0. 55 (b). P[z > z] = 0. 8708 => p[z z] = 0. 1292 (c). Problem b: in filling bags with flour, the automatic filling process dispenses amounts that follow a normal distribution with a mean of 10. 15 kg and a standard deviation of 0. 12 kg. The label on the bags indicates that the amount of flour in the bags is 10 kg. Define a random variable to be used in questions 1 to 3 below, and identify its probability distribution. Let x be the amount of flour in a randomly chosen bag; then. 0: bags that are overfilled are prone to bursting. Bags with fills of at least 10. 40 kg are prone to bursting.
