CQMS204-Statistics For Management Chapter6
Continuous Random Variable - Normal Distributions
• 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 )
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
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• 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.
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
b. higher = 47; lower = -100000000 = 0.91924
c. lower = 42; higher = 49 = 0.30864792
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
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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 )
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.
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;