BNAL 206 EXAM III QUESTIONS AND VERIFIED
ANSWERS
Continuous Probability Distributions - Answers - -Continuous random variable
-Continuous probability distribution
-The normal distribution
continuous random variable - Answers - -Values from interval of numbers
-absence of gaps
continuous probability distribution - Answers - -distribution of continuous random
variable
the normal distribution - Answers - -"Bell-shaped"
-Symmetrical
-Mean, Median, and Mode are equal
-Random variable has infinite range
Examples for Standardized Z - Answers - Look on z-table
P(Z < = 2.0) = 0.9772
P(Z > = 2.0) = 1 - P(Z<=2) = 1-0.9772 = 0.0228
P(-2 <= Z <= -1) = P(Z<=-2) - P(Z<=-1) = 0.1587 - 0.0228 = 0.1359
Standardized z-score formula - Answers - Z = (x-μ)/σ)
Standardizing example problem - Answers - A statistical analysis of 1,000 long-distance
telephone calls made from the headquarters of the Bricks and Clicks Computer
Corporation indicates that the length of these calls is normally distributed, with μ= 240
seconds and σ=40 seconds.
a.What is the probability that a call lasted less than 180 seconds?
Z = (X - μ)/σ
Z = (180 - 240)/40 = -1.5
P(X<180) = P(Z<-1.5) = 0.0668
b.What is the probability that a call lasted between 180 and 300 seconds?
Z = (180 - 240)/40 = -1.5
Z = (300 - 240)/40 = 1.5
P(180 < X < 300) = P(-1.5 < Z < 1.5) = P(Z<1.5) - P(Z<-1.5) = 0.9332 - 0.0668 = 0.8664
, Standardized z backwards - Answers - Look for closest value to given percentage value
in z-table values
P(Z<=?) = 0.60 Z = 0.25
P(Z>=?) = 0.40 = P(Z<=?) = 0.60 Z = 0.25
P(-? <=Z <=?) = 0.95 = P(Z<=?) = 0.025 Z = -1.96 and 1.96
Recovering X values for known probabilities Formula (normal distribution backwards) -
Answers - X = μ + Zσ
Normal distribution backwards example - Answers - A statistical analysis of 1,000 long-
distance telephone calls made from the headquarters of the Bricks and Clicks Computer
Corporation indicates that the length of these calls is normally distributed, with μ= 240
seconds and σ=40 seconds.
d.What is the length of a call if only 1% of all calls are shorter?
P(X<?) = 0.01
Z = -2.33
X = μ + Zσ
X = 240 - 2.33(40) = 146.8
What are sample statistics used to estimate? - Answers - Population parameters
Different samples provide different estimates - Answers - Large samples give better
estimates; but large samples cost more
How good is the estimate?
Sampling Distribution - Answers - -Theoretical probability distribution of a sample
statistic
-Sample mean, sample proportion
-Results from taking all possible samples of the same size
Standard error - Answers - the standard deviation of a sampling distribution for a large
population
central tendency - Answers - μx̅ = μ
Central Limit Theorem - Answers - As sample size gets large enough...the sampling
distribution becomes almost normal regardless of shape of population
How large is large enough for the Central Limit Theorem? - Answers - -For most
distributions, n >=30
-For fairly symmetric distributions, n >=15
-For normal distribution, the sampling distribution of the mean is always normally
distributed
ANSWERS
Continuous Probability Distributions - Answers - -Continuous random variable
-Continuous probability distribution
-The normal distribution
continuous random variable - Answers - -Values from interval of numbers
-absence of gaps
continuous probability distribution - Answers - -distribution of continuous random
variable
the normal distribution - Answers - -"Bell-shaped"
-Symmetrical
-Mean, Median, and Mode are equal
-Random variable has infinite range
Examples for Standardized Z - Answers - Look on z-table
P(Z < = 2.0) = 0.9772
P(Z > = 2.0) = 1 - P(Z<=2) = 1-0.9772 = 0.0228
P(-2 <= Z <= -1) = P(Z<=-2) - P(Z<=-1) = 0.1587 - 0.0228 = 0.1359
Standardized z-score formula - Answers - Z = (x-μ)/σ)
Standardizing example problem - Answers - A statistical analysis of 1,000 long-distance
telephone calls made from the headquarters of the Bricks and Clicks Computer
Corporation indicates that the length of these calls is normally distributed, with μ= 240
seconds and σ=40 seconds.
a.What is the probability that a call lasted less than 180 seconds?
Z = (X - μ)/σ
Z = (180 - 240)/40 = -1.5
P(X<180) = P(Z<-1.5) = 0.0668
b.What is the probability that a call lasted between 180 and 300 seconds?
Z = (180 - 240)/40 = -1.5
Z = (300 - 240)/40 = 1.5
P(180 < X < 300) = P(-1.5 < Z < 1.5) = P(Z<1.5) - P(Z<-1.5) = 0.9332 - 0.0668 = 0.8664
, Standardized z backwards - Answers - Look for closest value to given percentage value
in z-table values
P(Z<=?) = 0.60 Z = 0.25
P(Z>=?) = 0.40 = P(Z<=?) = 0.60 Z = 0.25
P(-? <=Z <=?) = 0.95 = P(Z<=?) = 0.025 Z = -1.96 and 1.96
Recovering X values for known probabilities Formula (normal distribution backwards) -
Answers - X = μ + Zσ
Normal distribution backwards example - Answers - A statistical analysis of 1,000 long-
distance telephone calls made from the headquarters of the Bricks and Clicks Computer
Corporation indicates that the length of these calls is normally distributed, with μ= 240
seconds and σ=40 seconds.
d.What is the length of a call if only 1% of all calls are shorter?
P(X<?) = 0.01
Z = -2.33
X = μ + Zσ
X = 240 - 2.33(40) = 146.8
What are sample statistics used to estimate? - Answers - Population parameters
Different samples provide different estimates - Answers - Large samples give better
estimates; but large samples cost more
How good is the estimate?
Sampling Distribution - Answers - -Theoretical probability distribution of a sample
statistic
-Sample mean, sample proportion
-Results from taking all possible samples of the same size
Standard error - Answers - the standard deviation of a sampling distribution for a large
population
central tendency - Answers - μx̅ = μ
Central Limit Theorem - Answers - As sample size gets large enough...the sampling
distribution becomes almost normal regardless of shape of population
How large is large enough for the Central Limit Theorem? - Answers - -For most
distributions, n >=30
-For fairly symmetric distributions, n >=15
-For normal distribution, the sampling distribution of the mean is always normally
distributed
