Unit 5 — Normal Distribution · Statistics. These are 10 questions from the Magmo3 bank on this lesson, from easy to hard. Solve each one, then open "Show answer and explanation". Magmo3 has 102 questions on this lesson, each with an explanation.
If the lengths of the students at a school follow a normal distribution whose mean is 160 cm and standard deviation is 4 cm, then the standard value Z corresponding to a length of 172 cm is ……
Correct answer: (A) 3
Explanation: Step 1: Z = (X − μ)/σ = (172 − 160)/4 Step 2: = 12/4 = 3 ∴ Z = 3 (12 is the difference before dividing by σ, and 43 comes from dividing 172 by 4).
If X is a normal random variable whose mean is 18 and standard deviation is 2.5, then P(X < 15) = P(Z < K) where K = ……
Correct answer: (B) −1.2
Explanation: Step 1: K = (X − μ)/σ = (15 − 18)/2.5 Step 2: = −3/2.5 = −1.2 ∴ K = −1.2 (−0.48 comes from dividing by the variance 6.25 instead of the standard deviation, and 6 from dividing 15 by 2.5).
The area of the region under the normal curve and above the X axis equals ……
Correct answer: (C) 1
Explanation: Step 1: One of the properties of the normal curve: the total area under the curve and above the X axis equals 1. ∴ The area = 1 (0.5 is the area of each of the two halves on either side of X = μ).
If Z is a standard normal random variable and P(0 ≤ Z ≤ 2) = 0.4772, then the probability that Z < −2 or Z > 2 equals ……
Correct answer: (D) 0.0456
Explanation: Step 1: P(−2 ≤ Z ≤ 2) = 2 × P(0 ≤ Z ≤ 2) = 2 × 0.4772 = 0.9544 Step 2: The required probability is the rest of the area: 1 − 0.9544 = 0.0456 ∴ The probability = 0.0456 (it covers both ends; each end alone is 0.0228).
If Z is a standard normal random variable and P(Z ≥ K) = 0.9616, then K = …… given that P(0 ≤ Z ≤ 1.77) = 0.4616 and P(0 ≤ Z ≤ 0.1) = 0.0398
Correct answer: (B) −1.77
Explanation: Step 1: The area is more than 0.5 and the inequality is "greater than", so K is negative. Step 2: P(K ≤ Z ≤ 0) = 0.9616 − 0.5 = 0.4616 = P(0 ≤ Z ≤ 1.77) ∴ K = −1.77 (1.77 forgets that K lies in the negative part).
If the weights of the students in a faculty follow a normal distribution whose mean is 68 kg and variance is 16 kg², then the probability that a randomly chosen student weighs more than 72 kg equals …… given that P(0 ≤ Z ≤ 1) = 0.3413 and P(0 ≤ Z ≤ 0.25) = 0.0987
Correct answer: (B) 0.1587
Explanation: Step 1: σ = √16 = 4, so Z = (72 − 68)/4 = 1 Step 2: P(X > 72) = P(Z > 1) = 0.5 − 0.3413 = 0.1587 ∴ The probability = 0.1587 (using the variance 16 instead of σ gives Z = 0.25 and the wrong answer 0.4013).
If X is a normal random variable and 95.44% of its values lie in the interval from μ − 2σ to μ + 2σ, which is the interval from 38 to 62, then P(X > 59) = …… given that P(0 ≤ Z ≤ 0.75) = 0.2734 and P(0 ≤ Z ≤ 1.5) = 0.4332
Correct answer: (A) 0.0668
Explanation: Step 1: The interval μ ± 2σ is [38, 62], so μ = (38 + 62)/2 = 50 and 4σ = 24, so σ = 6 Step 2: P(X > 59) = P(Z > 1.5) = 0.5 − 0.4332 = 0.0668 ∴ P(X > 59) = 0.0668 (0.2266 comes from taking 62 − 38 = 2σ, so σ = 12; 0.0122 from taking it as 6σ, so σ = 4; 0.4332 from forgetting to subtract from 0.5).
If X is a normal random variable whose mean is μ and standard deviation is σ, and two of its values are chosen at random independently, then the probability that one of them is greater than μ + σ and the other is less than μ − σ, to four decimal places, = …… given that P(0 ≤ Z ≤ 1) = 0.3413
Correct answer: (A) 0.0504
Explanation: Step 1: P(X > μ + σ) = P(Z > 1) = 0.5 − 0.3413 = 0.1587, and from symmetry P(X < μ − σ) = 0.1587 Step 2: The two values are chosen independently, and the event happens if the first is large and the second small, or the other way round Step 3: 2 × 0.1587 × 0.1587 ≈ 0.0504 ∴ The probability ≈ 0.0504 (0.0252 ignores the second order, and 0.3174 adds instead of multiplying).
If X is a normal random variable whose mean is μ and standard deviation is σ, and P(X > μ + 5) = 0.1056, then the probability that X differs from μ by more than 6 = …… given that P(0 ≤ Z ≤ 1.25) = 0.3944 and P(0 ≤ Z ≤ 1.5) = 0.4332
Correct answer: (A) 0.1336
Explanation: Step 1: P(Z > 5/σ) = 0.1056, so P(0 ≤ Z ≤ 5/σ) = 0.3944, 5/σ = 1.25 and σ = 4 Step 2: P(|X − μ| > 6) = P(Z < −1.5) + P(Z > 1.5) = 2 × (0.5 − 0.4332) = 0.1336 ∴ The probability = 0.1336 (0.0668 is one end only, 0.8664 is the probability of differing by less than 6, and 0.2112 = 2 × 0.1056 uses 5 instead of 6).
If X is a normal random variable whose mean is 70 and P(X > 78) = 0.0548, then P(X < 65) = …… given that P(0 ≤ Z ≤ 1.6) = 0.4452 and P(0 ≤ Z ≤ 1) = 0.3413
Correct answer: (C) 0.1587
Explanation: Step 1: 0.5 − 0.0548 = 0.4452, so (78 − 70)/σ = 1.6 and σ = 5 Step 2: P(X < 65) = P(Z < (65 − 70)/5) = P(Z < −1) = 0.5 − 0.3413 = 0.1587 ∴ P(X < 65) = 0.1587 (0.0548 wrongly assumes that 65 and 78 are symmetric about the mean).
The first 10 questions in every subject are free, each with an explanation.