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5-2 Some Practical Applications of the Normal Distribution: practice questions

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 74 questions on this lesson, each with an explanation.

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Question 1Easy

A machine produces cylinders whose lengths follow a normal distribution whose mean is 56 cm and standard deviation is 2 cm. A cylinder is valid if its length is between 51 cm and 60 cm. In terms of the standard normal variable Z, the probability that a cylinder is valid is ……

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Correct answer: (D) P(−2.5 < Z < 2)

Explanation: Step 1: Z = (51 − 56)/2 = −2.5 and Z = (60 − 56)/2 = 2 Step 2: P(51 < X < 60) = P(−2.5 < Z < 2) ∴ The probability = P(−2.5 < Z < 2) (−5 and 4 are the differences before dividing by σ).

Question 2Easy

If the lengths of 2000 students at a faculty follow a normal distribution whose mean is 170 cm, then the expected number of students whose lengths are more than 170 cm = ……

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Correct answer: (A) 1000

Explanation: Step 1: 170 is the mean, and the straight line X = μ divides the area under the curve into two regions, each = 0.5 Step 2: Number = 2000 × 0.5 = 1000 ∴ 1000 students

Question 3Easy

If the values of a phenomenon follow a normal distribution, and the area from μ − 3σ to μ + 3σ is 99.74% of the total area, then the percentage of the values outside this interval = ……

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Correct answer: (B) 0.26%

Explanation: Step 1: Percentage outside = 100% − 99.74% = 0.26% ∴ 0.26% (half of it, 0.13%, is in each end; 4.56% is the percentage outside μ − 2σ to μ + 2σ).

Question 4Medium

If the weights of the students at a faculty follow a normal distribution whose mean is 68.5 kg and standard deviation is 2.5 kg, then the percentage of the students whose weights are between 67.5 kg and 71 kg = …… given that P(0 ≤ Z ≤ 0.4) = 0.1554 and P(0 ≤ Z ≤ 1) = 0.3413

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Correct answer: (B) 49.67%

Explanation: Step 1: Z = (67.5 − 68.5)/2.5 = −0.4 and Z = (71 − 68.5)/2.5 = 1 Step 2: P(−0.4 < Z < 1) = 0.1554 + 0.3413 = 0.4967 ∴ The percentage = 49.67% (18.59% comes from subtracting the two areas).

Question 5Medium

If the lengths of the students at a high school follow a normal distribution whose mean is 160 cm and standard deviation is 5 cm, then the probability that a student's length differs from the mean by more than 8 cm equals …… given that P(0 ≤ Z ≤ 1.6) = 0.4452

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Correct answer: (B) 0.1096

Explanation: Step 1: P(|X − 160| > 8) = 1 − P(152 < X < 168) = 1 − P(−1.6 < Z < 1.6) Step 2: = 1 − 2 × 0.4452 = 0.1096 ∴ The probability = 0.1096 (0.8904 is the probability of differing by not more than 8 cm, and 0.0548 is one end only).

Question 6Medium

If the salary of 1000 families in a city is a normal random variable whose mean is 170 LE and standard deviation is 20 LE, then the expected number of families whose salaries are more than 150 LE, to the nearest whole number, = …… given that P(0 ≤ Z ≤ 1) = 0.3413

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Correct answer: (A) 841

Explanation: Step 1: Z = (150 − 170)/20 = −1, so P(Z > −1) = 0.5 + 0.3413 = 0.8413 Step 2: Number = 1000 × 0.8413 = 841.3 ≈ 841 ∴ About 841 families (159 is the number of families whose salaries are less than 150 LE).

Question 7Medium

In an exam taken by 10000 students, the marks have a normal distribution whose mean is 70 and standard deviation is 5. The expected number of students whose marks are more than 78 = …… given that P(0 ≤ Z ≤ 1.6) = 0.4452

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Correct answer: (B) 548

Explanation: Step 1: Z = (78 − 70)/5 = 1.6, so P(Z > 1.6) = 0.5 − 0.4452 = 0.0548 Step 2: Number = 10000 × 0.0548 = 548 ∴ 548 students (1096 counts both ends).

Question 8Hard

Two machines each produce lengths with mean 50 mm, and the lengths from each follow a normal distribution. 4.56% of the first machine's lengths lie outside the interval from 46 to 54 mm, and 31.74% of the second machine's lengths lie outside the interval from 47 to 53 mm. Which statement is true? Given that P(0 ≤ Z ≤ 1) = 0.3413 and P(0 ≤ Z ≤ 2) = 0.4772

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Correct answer: (D) σ is 2 for the first and 3 for the second, so the first is more consistent

Explanation: Step 1: First: each end is 2.28%, so P(0 ≤ Z ≤ a) = 0.4772, 4/σ = 2 and σ = 2 Step 2: Second: each end is 15.87%, so P(0 ≤ Z ≤ a) = 0.3413, 3/σ = 1 and σ = 3 Step 3: A smaller standard deviation means less spread about the mean ∴ The first is more consistent (σ = 4 comes from taking 46 to 54 as μ ± σ).

Question 9Hard

In an exam the marks follow a normal distribution. If 15.87% of the students got more than 80 marks and 2.28% of them got less than 50 marks, then the mean of the marks μ = …… given that P(0 ≤ Z ≤ 1) = 0.3413 and P(0 ≤ Z ≤ 2) = 0.4772

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Correct answer: (D) 70

Explanation: Step 1: 0.5 − 0.1587 = 0.3413, so (80 − μ)/σ = 1; 0.5 − 0.0228 = 0.4772, so (50 − μ)/σ = −2 Step 2: Subtracting: 30 = 3σ, so σ = 10 Step 3: μ = 80 − 10 = 70 ∴ μ = 70 (65, the middle of 50 and 80, is wrong because the two percentages are not equal, and 10 is σ).

Question 10Hard

The lengths of a certain type of plants follow a normal distribution whose mean is 50 cm and standard deviation is σ. If the lengths of 10.56% of these plants are less than 45 cm, then the variance of the lengths of these plants = …… given that P(0 ≤ Z ≤ 1.25) = 0.3944

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Correct answer: (B) 16

Explanation: Step 1: P(X < 45) = 0.1056, and 0.5 − 0.1056 = 0.3944, so Z = −1.25 Step 2: (45 − 50)/σ = −1.25, so σ = 4 Step 3: The variance σ² = 4² = 16 ∴ The variance = 16 (4 is the standard deviation, not the variance).

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