Textbook Notes (368,786)
QMS 230 (8)
Chapter 6

Chapter 6 Worksheet

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School
Department
Quantitative Methods
Course
QMS 230
Professor
Douglas Mc Kessock
Semester
Fall

Description
CQMS204-Statistics For Management Chapter6 Chapter 6 Continuous Random Variable - Normal Distributions Outcomes: • Reveal the characteristics of the Normal Distribution • Recognize the Standard Normal Distribution • Calculate the normal probabilities Characteristic of Normal Distribution (We use x, y values) • bell-shaped and has a single peak. (mode, mean, median is the highest point) • is symmetric about the mean. • is asymptotic, that is, the curve gets closer and closer to the x-axis but never actually touches it. (gets very close to a-axis but never touches) • the total area under the curve is 1 • is denoted by N(μ ,σ ) (z value) Negative value (Mean = 0) Positive value  The Standard Normal Distribution (We use z-value, z-score values) • With a mean of 0 and a standard deviation of 1, N(0, 1). • Z-value or Z-score – the units marked on the horizontal axis of the standard normal curve and are called z values or z scores. • the z values on the right side of the mean are positive Winter2011 Page #1 CQMS204-Statistics For Management Chapter6 • the z values on the left side of the mean are negative • the standard normal distribution table Standardizing a Normal Distribution, Z • If the random variable X has a normal distribution with mean μ and standard deviation σ , that are different from 0 and 1 respectively. • Convert the given normal distribution to the standard normal distribution • Convert the random variable X to the standard normal variable Z, using the equation x − μ z = σ • Thus, P(X < x) = P(Z < z), where z is called the z-value. The process to change it from x value to z value is called transformation. Example1 Let x be a normal random variable with its mean equal to 40 and standard deviation equal to 5. Determine the following probabilities: (x ≥ 50) (x < 47) (42 ≤ x ≤ 49) a. P b. P c. P In continuous, P(x > 50) = P(x >= 50) Lower = 50 (x >= 50)  min Upper = infinite (choose the num but put lots of zeros) 100000000 (x > 50)  max Standard Dev. = 5 Mean = 40 a. 0.02275013 b. higher = 47; lower = -100000000 = 0.91924 c. lower = 42; higher = 49 = 0.30864792 Example2 Let x be a normal random variable with its mean equal to 12 and standard deviation equal to 2. Determine the following probabilities: a. P (x ≥ 12) b. P (x < 12 .) c. P (9.24 ≤ x ≤ 11 .) M = 12 Standard Dev. = 2 a. lower = 12; higher = 10000000000; = 0.5 b. lower = -1000000000; higher = 12.8; = 0.65542174 c. lower = 9.24; higher = 11.8; = 0.37637884 Winter2011 Page #2 CQMS204-Statistics For Management Chapter6 Example3 a. What is the area under the normal curve between z = –2.34 and z = –1.45? P(-2.34 < Z < -1.45) (For z, mean is always = 0, and stand. Dev. = 1) = 0.0638 b. What is the area under the normal curve between z = -9.0 and z = 0? P(-9.0 < Z < 0) = 0.5 c. Determine i. P (−1.45 ≤ z ≤ 2.06 ) ii. (z ≥ −1.47 ) i. 0.90677 ii. 0.929 Example4 The monthly credit card bills for households in Toronto are normally distributed, with mean of \$1476.30 and the standard deviation of \$385.89. a. Define the random variable X b. Determine the probability that a randomly selected bill is i. less than \$1600 ii. more than \$1200 iii. between \$1100 and \$1700 M = 1476.30 Standard Dev. = 385.89 a. Monthly credit card bills for households in Toronto. b. Answers: i. P(x < 1600) Lower = -1000000; Higher = 1600 a. = 0.6257 ii. P(x > 1200) Lower = 1200; Higher = 10000000 a. = 0.763 iii. P(1100 < x < 1700) Lower = 1100;
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