BSTAT EXAM 6 QUESTIONS AND 100%
CORRECT ANSWERS!!
Let X be normally distributed with mean µ = 25 and standard deviation σ = 5. Find the
value x such that P(X ≥ x) = 0.1736.
29.70
The appropriate Excel function is =NORM.INV(1-0.1736,25,5) =29.70
Find the probability P(−1.96 ≤ Z ≤ 0).
0.4750
Compute P(z1 ≤ Z ≤ z2) = P(Z ≤ z2) − P(Z ≤ z1).
Consider the following cumulative distribution function for the discrete random variable
X.
x - P(X ≤ x)
1 - 0.10
2 - 0.35
3 - 0.75
4 - 0.85
5 - 1.00
What is the probability that X is greater than 3?
0.25
P(X ≤ 3) = 1 −P(X ≤ 3) = 1 − 0.75 = 0.25
For a particular clothing store, a marketing firm finds that 16% of $10-off coupons
delivered by mail are redeemed. Suppose six customers are randomly selected and are
mailed $10-off coupons. What is the expected number of coupons that will be redeemed?
, 0.96
The expected value of a binomial random variable is calculated as E(X) = μ = np.
μ = 6 × 0.16 = 0.96
Suppose the average price of gasoline for a city in the United States follows a continuous
uniform distribution with a lower bound of $3.50 per gallon and an upper bound of $3.80
per gallon. What is the probability a randomly chosen gas station charges more than $3.70
per gallon?
0.3333
=(3.8 − 3.7)/(3.8 − 3.5) =0.3333
Consider the following probability distribution.
xi - P(X = xi)
0 - 0.1
1 - 0.2
2 - 0.4
3 - 0.3
The expected value is _____.
1.90
The variance of the discrete random variable X is calculated as
Var(X) = σ^2 = ∑(xi - μ)^2 P(X = xi).
E(X) = 0 × 0.10 + 1 × 0.20 + 2 × 0.40 + 3 × 0.30 = 1.9
Var(X) = (0 - 1.9)^2 × 0.10 + (1 - 1.9)^2 × 0.20 + (2 - 1.9)^2 × 0.40 + (3 - 1.9)^2 × 0.30 = 0.89
Find the z value such that P(−z ≤ Z ≤ z) = 0.95.
z = 1.96
CORRECT ANSWERS!!
Let X be normally distributed with mean µ = 25 and standard deviation σ = 5. Find the
value x such that P(X ≥ x) = 0.1736.
29.70
The appropriate Excel function is =NORM.INV(1-0.1736,25,5) =29.70
Find the probability P(−1.96 ≤ Z ≤ 0).
0.4750
Compute P(z1 ≤ Z ≤ z2) = P(Z ≤ z2) − P(Z ≤ z1).
Consider the following cumulative distribution function for the discrete random variable
X.
x - P(X ≤ x)
1 - 0.10
2 - 0.35
3 - 0.75
4 - 0.85
5 - 1.00
What is the probability that X is greater than 3?
0.25
P(X ≤ 3) = 1 −P(X ≤ 3) = 1 − 0.75 = 0.25
For a particular clothing store, a marketing firm finds that 16% of $10-off coupons
delivered by mail are redeemed. Suppose six customers are randomly selected and are
mailed $10-off coupons. What is the expected number of coupons that will be redeemed?
, 0.96
The expected value of a binomial random variable is calculated as E(X) = μ = np.
μ = 6 × 0.16 = 0.96
Suppose the average price of gasoline for a city in the United States follows a continuous
uniform distribution with a lower bound of $3.50 per gallon and an upper bound of $3.80
per gallon. What is the probability a randomly chosen gas station charges more than $3.70
per gallon?
0.3333
=(3.8 − 3.7)/(3.8 − 3.5) =0.3333
Consider the following probability distribution.
xi - P(X = xi)
0 - 0.1
1 - 0.2
2 - 0.4
3 - 0.3
The expected value is _____.
1.90
The variance of the discrete random variable X is calculated as
Var(X) = σ^2 = ∑(xi - μ)^2 P(X = xi).
E(X) = 0 × 0.10 + 1 × 0.20 + 2 × 0.40 + 3 × 0.30 = 1.9
Var(X) = (0 - 1.9)^2 × 0.10 + (1 - 1.9)^2 × 0.20 + (2 - 1.9)^2 × 0.40 + (3 - 1.9)^2 × 0.30 = 0.89
Find the z value such that P(−z ≤ Z ≤ z) = 0.95.
z = 1.96
