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PORTAGE LEARNING STATISTICS MODULE 4 THE Z-TEST
QUESTIONS AND ANSWERS WITH COMPLETE SOLUTIONS |
NEW UPDATE 2025
## SECTION 1: BINOMIAL DISTRIBUTION
Question 1
A factory has eight safety systems. During an emergency, the probability of any one of the safety systems failing is 0.08. What is the probability that six or more safety systems will fail during an emergency? A. 6.36 × 10⁻⁶ B. 5.21 × 10⁻⁵ C. 1.54 × 10⁻⁷ D. 2.83 × 10⁻⁴ Answer: A. 6.36 × 10⁻⁶
Rationale: This is a binomial probability problem because we have a fixed
number of trials (n=8), two possible outcomes (fail or not fail), constant probability of failure (p=0.08), and independent events. We need P(x≥6) = P(x=6) + P(x=7) + P(x=8). The binomial probability formula is: f(x) = [n! / (x!(n-x)!)] × p^x × (1-p)^(n-x) Calculations:
- For x=6: [8!/(6!2!)] × (0.08⁶) × (0.92²) = 6.21 × 10 ⁻⁶
- For x=7: [8!/(7!1!)] × (0.08⁷) × (0.92¹) = 1.54 × 10⁻⁷
- For x=8: [8!/(8!0!)] × (0.08⁸) × (0.92⁰) = 1.68 × 10 ⁻⁹ Sum = 6.21×10⁻⁶ + 1.54× 10 ⁻⁷ + 1.68× 10 ⁻⁹ = 6.36 × 10 ⁻⁶
Question 2
In a large shipment of clocks, it has been discovered that 15% of the clocks are defective. Suppose you choose 9 clocks at random. What is the probability that exactly 2 of the clocks are defective? A. 0. B. 0. C. 0. D. 0. Answer: A. 0.2597
Rationale: This is a binomial probability problem with:
- n = 9 (number of trials)
- x = 2 (number of successes/defective clocks)
- p = 0.15 (probability of a defective clock)
- q = 1 - p = 0.85 (probability of non-defective) Apply the binomial formula: f(2) = [9!/(2! × 7!)] × (0.15²) × (0.85⁷) = 0.2597
Question 3
An archer is shooting arrows at a target. She hits the target 77% of the time. If she takes 20 shots at the target, what is the probability that she will hit the target exactly 15 times?
- The probability of success (p) is the same for each trial
- All trials are independent The probability of success must remain constant across all trials—it cannot change.
Question 5
A binomial experiment has n=14 trials and p=0.55. Which expression correctly represents the probability of exactly 7 successes? A. [14!/(7!×7!)] × (0.55⁷) × (0.45⁷) B. [14!/(7!×7!)] × (0.55⁷) × (0.55⁷) C. [14!/(7!×6!)] × (0.55⁷) × (0.45⁷) D. 14 × 0.55⁷ × 0.45⁷ Answer: A. [14!/(7!×7!)] × (0.55⁷) × (0.45⁷)
Rationale: The binomial formula is:
f(x) = [n!/(x!(n-x)!)] × p^x × (1-p)^(n-x) For n=14, x=7, p=0.55:
- The combination term: 14!/(7! × 7!)
- Success term: 0.55⁷
- Failure term: (0.45)⁷ (since 1-0.55 = 0.45)
SECTION 2: NORMAL DISTRIBUTION AND Z-SCORES
Question 6
A company manufactures rods. The lengths are normally distributed with mean 3. inches and standard deviation 0.35 inches. If you choose a rod at random, what is the probability it will be less than 3.5 inches? A. 0. B. 0. C. 0. D. 0. Answer: B. 0.2843
Rationale: For a normal distribution, we calculate the z-score first:
z = (x - μ) / σ = (3.5 - 3.7) / 0.35 = -0.2 / 0.35 = -0. P(Z ≤ -0.57) = 0.2843 from the standard normal table. This means approximately 28.43% of rods are less than 3.5 inches.
Question 7
Using the same rod distribution (μ=3.7, σ=0.35), what is the probability that a randomly chosen rod will be greater than 3.5 inches? A. 0. B. 0. C. 0. D. 0. Answer: C. 0.7157
A. 0.
B. 0.
C. 0.
D. 0.
Answer: B. 0.5793
Rationale: First calculate z-score:
z = (4.5 - 4.4)/0.5 = 0.1/0.5 = 0. P(Z ≤ 0.2) = 0.5793 (from standard normal table)
Question 10
Using μ=4.4, σ=0.5, find P(x ≥ 4.0). A. 0. B. 0. C. 0. D. 0. Answer: B. 0.7881
Rationale: Calculate z = (4.0 - 4.4)/0.5 = -0.4/0.5 = -0.
P(Z ≥ -0.8) = 1 - P(Z ≤ -0.8) = 1 - 0.2119 = 0.7881
Question 11
For μ=4.4, σ=0.5, find P(3.8 ≤ x ≤ 4.7). A. 0. B. 0. C. 0. D. 0. Answer: A. 0.6107
Rationale: Calculate z-scores:
- z₁ = (3.8 - 4.4)/0.5 = -0.6/0.5 = -1.
- z₂ = (4.7 - 4.4)/0.5 = 0.3/0.5 = 0. P(-1.2 ≤ Z ≤ 0.6) = P(Z ≤ 0.6) - P(Z ≤ -1.2) = 0.7257 - 0.1151 = 0.6107
Question 12
What z-score corresponds to the 90th percentile of the standard normal distribution? A. 1. B. 1. C. 1. D. 2. Answer: A. 1.28
Answer: A. 0.6826
Rationale: z₁ = (62-65)/3 = -1.0; z₂ = (68-65)/3 = 1.
P(-1.0 ≤ Z ≤ 1.0) = 0.8413 - 0.1587 = 0.6826
Question 15
Which z-score would be considered an "unusual" observation? A. z = 1. B. z = -1. C. z = 2. D. z = 0. Answer: C. z = 2.3
Rationale: Values with |z| > 2 are generally considered unusual because they
occur less than 5% of the time. z = 2.3 corresponds to approximately the 99th percentile, meaning only about 1% of observations are more extreme.
SECTION 3: EXPECTED VALUE AND VARIANCE
Question 16
A life insurance salesperson expects to sell between zero and five policies per day. The probability distribution is given:
Policies Sold (x)Probability f(x)
What is the expected number of policies sold per day? A. 2. B. 2. C. 3. D. 2. Answer: B. 2.96
Rationale: Expected value E(x) = μ = Σ[x × f(x)]
= (0×0.04) + (1×0.11) + (2×0.23) + (3×0.26) + (4×0.19) + (5×0.17)
= 0 + 0.11 + 0.46 + 0.78 + 0.76 + 0.85 = 2.96 policies per day
Question 17
For the same probability distribution, what is the variance? A. 1. B. 1. C. 2. D. 1.
What formula is used to calculate the expected value of a discrete random variable? A. E(x) = Σx B. E(x) = Σ[x + f(x)] C. E(x) = Σ[x × f(x)] D. E(x) = Σ[f(x)] Answer: C. E(x) = Σ[x × f(x)]
Rationale: The expected value (mean) of a discrete random variable is the
sum of each possible value multiplied by its probability: E(x) = Σ[x × P(x)]
Question 20
For a discrete random variable, the sum of all probabilities must equal: A. 0 B. 0. C. 1 D. 100 Answer: C. 1
Rationale: The total probability for all possible outcomes in a probability
distribution must sum to exactly 1, representing 100% probability.
SECTION 4: STANDARD NORMAL PROBABILITIES
Question 21
Find P(Z ≤ 1.27). A. 0. B. 0. C. 0. D. 0. Answer: B. 0.8980
Rationale: Look up z=1.27 in the standard normal table. The value 1.
corresponds to 1.2 in the row and 0.07 in the column. P(Z ≤ 1.27) = 0.8980 (approximately 0.89796)
Question 22
Find P(Z ≥ -0.73). A. 0. B. 0. C. 0. D. 0. Answer: C. 0.7673
Rationale: P(Z ≥ -0.73) = 1 - P(Z ≤ -0.73). From the table, P(Z ≤ -0.73) =
0.2327. Therefore, 1 - 0.2327 = 0.7673
Question 25
Find P(Z ≥ -0.34). A. 0. B. 0. C. 0. D. 0. Answer: C. 0.6331
Rationale: P(Z ≥ -0.34) = 1 - P(Z ≤ -0.34) = 1 - 0.3669 = 0.6331
Question 26
Find P(-1.14 ≤ Z ≤ 0.55). A. 0. B. 0. C. 0. D. 0. Answer: A. 0.7088
Rationale: P(-1.14 ≤ Z ≤ 0.55) = P(Z ≤ 0.55) - P(Z ≤ -1.14)
- P(Z ≤ 0.55) = 0.
- P(Z ≤ -1.14) = 0.
- Difference = 0.7088 - 0.1271 = 0.5817
Question 27
Find P(Z ≤ -0.86). A. 0. B. 0. C. 0. D. 0. Answer: C. 0.1949
Rationale: From the standard normal table, z=-0.86 corresponds to the
intersection of row -0.8 and column 0.06: P(Z ≤ -0.86) = 0.1949
Question 28
Find P(Z ≤ 0.86). A. 0. B. 0. C. 0. D. 0. Answer: A. 0.8051
Rationale: The 75th percentile means 75% of observations fall below this
value. From the standard normal table, z = 0.67 gives approximately 0.7486, and z = 0.68 gives 0.7517. Interpolating gives z ≈ 0.67.
SECTION 5: PROBABILITY AND PERCENTILES
Question 31
What is the probability that a standard normal random variable Z is less than 0? A. 0. B. 0. C. 0. D. 1. Answer: B. 0.50
Rationale: The standard normal distribution is symmetric about the mean of
- Therefore, exactly half of the values fall below the mean: P(Z < 0) = 0.50.
Question 32
What is the probability that a standard normal random variable Z is between -1 and 1? A. 0. B. 0. C. 0. D. 0.
Answer: B. 0.6826
Rationale: P(-1 ≤ Z ≤ 1) = P(Z ≤ 1) - P(Z ≤ -1) = 0.8413 - 0.1587 =
0.6826. Approximately 68% of values fall within 1 standard deviation of the mean.
Question 33
What is the probability that a standard normal random variable Z is between -2 and 2? A. 0. B. 0. C. 0. D. 0. Answer: B. 0.9544
Rationale: Approximately 95% of values in a normal distribution fall within 2
standard deviations of the mean: 0.9544
Question 34
For a normally distributed variable, approximately what percentage of observations fall within 3 standard deviations of the mean? A. 68% B. 95%
