Normal Distribution MCQs

1. If a normally distributed variable has a mean of 50 and a standard deviation of 10, what is the probability that a randomly selected value is less than 60?

a) 0.8413 b) 0.1587 c) 0.6826 d) 0.5000

Correct Answer: a) 0.8413

Explanation:
To find P(X < 60) where X ~ N(50, 10), we need to standardize: Z = (X - μ)/σ = (60 - 50)/10 = 1. From the standard normal table, P(Z < 1) = 0.8413, which means there's an 84.13% probability that a randomly selected value is less than 60.

2. Which of the following is true about the standard normal distribution?

a) The mean is 1 and the standard deviation is 0. b) The total area under the curve is 1. c) It is skewed to the right. d) The curve is rectangular in shape.

Correct Answer: b) The total area under the curve is 1.

Explanation:
The standard normal distribution has mean 0 and standard deviation 1 (not mean 1 and SD 0). It is symmetric (not skewed) and bell-shaped (not rectangular). However, like all probability distributions, the total area under the curve equals 1, representing the total probability.

3. In a normal distribution, approximately what percentage of data falls within ±2 standard deviations of the mean?

a) 50% b) 68% c) 95% d) 99.7%

Correct Answer: c) 95%

Explanation:
According to the empirical rule for normal distributions: approximately 68% of data falls within ±1 standard deviation, 95% within ±2 standard deviations, and 99.7% within ±3 standard deviations of the mean.

4. If X ~ N(100, 25), what is the Z-score for X = 115?

a) 1.5 b) 2.0 c) 3.0 d) 0.5

Correct Answer: c) 3.0

Explanation:
For X ~ N(100, 25), note that σ = 5 (since variance = 25, standard deviation = √25 = 5). The Z-score is calculated as Z = (X - μ)/σ = (115 - 100)/5 = 15/5 = 3.0.

5. If P(X < k) = 0.9772 in N(50, 10), what is k?

a) 60 b) 70 c) 80 d) 90

Correct Answer: b) 70

Explanation:
Given P(X < k) = 0.9772, we find the corresponding Z-score = 2 (as P(Z < 2) = 0.9772). Then using Z = (k - μ)/σ, we get 2 = (k - 50)/10, so k = 50 + 2(10) = 70.

6. What shape does a normal distribution curve have?

a) Rectangular b) Bell-shaped c) U-shaped d) Linear

Correct Answer: b) Bell-shaped

Explanation:
The normal distribution has a characteristic symmetric bell shape, with the highest point at the mean and tapering off symmetrically on both sides. This is one of the defining visual characteristics of normal distributions.

7. The area under the standard normal distribution between Z = -1 and Z = 1 is approximately:

a) 68.26% b) 95.44% c) 34.13% d) 99.73%

Correct Answer: a) 68.26%

Explanation:
According to the empirical rule, approximately 68.26% of data falls within ±1 standard deviation of the mean in a normal distribution. Since the standard normal distribution has mean 0 and standard deviation 1, the area between Z = -1 and Z = 1 is 68.26%.

8. If Z is a standard normal variable, what is P(Z > 0)?

a) 0.6826 b) 0.5 c) 0.8413 d) 1.0

Correct Answer: b) 0.5

Explanation:
Since the standard normal distribution is symmetric around 0, exactly half of the area under the curve lies to the right of 0. Therefore, P(Z > 0) = 0.5 or 50%.

9. The probability that a patient recovers from a delicate heart operation is 0.9. Out of the next 100 patients, what is the probability that between 84 and 95 (inclusive) survive?

a) 0.9999 b) 0.2323 c) 0.9514 d) 0.6768

Correct Answer: c) 0.9514

Explanation:
This is a normal approximation to the binomial. With n=100, p=0.9, we have μ=np=90 and σ=√(np(1-p))=√(100×0.9×0.1)=3. P(84 ≤ X ≤ 95) = P((83.5-90)/3 ≤ Z ≤ (95.5-90)/3) = P(-2.17 ≤ Z ≤ 1.83) = P(Z ≤ 1.83) - P(Z ≤ -2.17) = 0.9664 - 0.0150 = 0.9514.

10. A coin is tossed 400 times. Use the normal approximation to find the probability of obtaining exactly 205 heads.

a) 0.0352 b) 0.0101 c) 0.9898 d) 0.5650

Correct Answer: a) 0.0352

Explanation:
For a binomial approximation, we use continuity correction. With n=400, p=0.5, μ=200, σ=10. P(X=205) ≈ P(204.5 < X < 205.5) = P(0.45 < Z < 0.55) = P(Z < 0.55) - P(Z < 0.45) ≈ 0.7088 - 0.6736 = 0.0352.

11. A multiple-choice quiz has 200 questions, each with 4 options. What is the probability of getting 25 to 30 correct answers by guessing for 80 unknown questions?

a) 0.9998 b) 0.7776 c) 0.1196 d) 0.1234

Correct Answer: c) 0.1196

Explanation:
With n=80 questions, p=1/4 for guessing, μ=np=20, σ=√(np(1-p))=√(80×0.25×0.75)=3.87. P(25 ≤ X ≤ 30) = P((24.5-20)/3.87 ≤ Z ≤ (30.5-20)/3.87) = P(1.16 ≤ Z ≤ 2.71) = P(Z ≤ 2.71) - P(Z ≤ 1.16) = 0.9966 - 0.8770 = 0.1196.

12. Which theorem justifies the use of normal approximation in large sample binomial problems?

a) Central Limit Theorem b) Law of Large Numbers c) Bayes' Theorem d) Chebyshev's Theorem

Correct Answer: a) Central Limit Theorem

Explanation:
The Central Limit Theorem states that the sum (or average) of a large number of independent, identically distributed random variables approaches a normal distribution regardless of the original distribution. This allows us to use normal approximation for binomial distributions when n is large.

13. What continuity correction is used when approximating binomial with normal?

a) Add 1 to X b) Subtract 1 from X c) Use X ± 0.5 d) Multiply X by 0.5

Correct Answer: c) Use X ± 0.5

Explanation:
When approximating discrete distributions (like binomial) with continuous distributions (like normal), we use a continuity correction of ±0.5. For example, P(X = k) is approximated as P(k-0.5 < X < k+0.5).

14. What is the MGF of a normal distribution N(μ, σ²)?

a) e^(μt + σ²t²/2) b) e^(μt - σ²t²/2) c) μt + σ²t²/2 d) e^(t²)

Correct Answer: a) e^(μt + σ²t²/2)

Explanation:
The moment generating function (MGF) for a normal distribution X ~ N(μ, σ²) is M_X(t) = exp(μt + σ²t²/2). This formula uniquely characterizes the normal distribution and is used to derive its moments.

15. Which of the following properties is true for the MGF of a normal distribution?

a) It exists only for t=0 b) The nth moment is obtained by the nth derivative of the MGF at t = 0 c) It is only defined for discrete distributions d) It is not differentiable

Correct Answer: b) The nth moment is obtained by the nth derivative of the MGF at t = 0

Explanation:
One of the key properties of moment generating functions is that the nth moment E[X^n] can be obtained by taking the nth derivative of the MGF with respect to t and evaluating it at t = 0. This is true for all distributions with valid MGFs, including the normal distribution.

16. The MGF of a standard normal variable Z ~ N(0, 1) is:

a) e^(t²) b) e^(t²/2) c) e^t d) t²

Correct Answer: b) e^(t²/2)

Explanation:
For a standard normal distribution Z ~ N(0, 1), we substitute μ = 0 and σ² = 1 into the general normal MGF formula e^(μt + σ²t²/2) to get e^(0*t + 1*t²/2) = e^(t²/2).

17. If two random variables have the same MGF, what can be concluded?

a) They are independent b) They have the same mean only c) They have the same distribution d) One must be normal

Correct Answer: c) They have the same distribution

Explanation:
The moment generating function uniquely determines a probability distribution. This means that if two random variables have the same MGF, they must have the same probability distribution, which implies they have the same moments, including mean, variance, etc.

18. Which moment does the second derivative of MGF at t=0 represent?

a) Mean b) Variance c) Second raw moment d) Standard deviation

Correct Answer: c) Second raw moment

Explanation:
The nth derivative of the MGF evaluated at t=0 gives the nth raw moment, E[X^n]. So the second derivative evaluated at t=0 gives E[X²], which is the second raw moment. This is not the same as variance, which is E[(X-μ)²].

19. If M_X(t) = e^(2t + 3t²), then the distribution of X is:

a) N(2, 9) b) N(3, 2) c) N(2, 3) d) N(3, 9)

Correct Answer: c) N(2, 3)

Explanation:
Comparing the given MGF M_X(t) = e^(2t + 3t²) with the normal MGF form e^(μt + σ²t²/2), we can identify that μ = 2 and σ²t²/2 = 3t², which means σ² = 6 (since t² coefficients must match). Therefore, X ~ N(2, 6). [Note: There seems to be a discrepancy in the options; N(2, 3) should actually be N(2, 6)]

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