CFA考试《定量方法》习题练习：标准正态分布表

来源：正保会计网校 2020-09-08

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1、The following is an excerpt from the cumulative distribution function for the

standard normal random variable table:

Cumulative Probabilities for a Standard Normal Distribution

P(Z ≤ x) = N(x) for x ≥ 0 or P(Z ≤ z) = N(z) for z ≥ 0

or z 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09

-----------------------------------------------------------------------------

0.10 0.5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636 0.5675 0.5714 0.5753

0.20 0.5793 0.5832 0.5871 0.5910 0.5948 0.5987 0.6026 0.6064 0.6103 0.6141

0.30 0.6179 0.6217 0.6255 0.6293 0.6331 0.6368 0.6406 0.6443 0.6480 0.6517

0.40 0.6554 0.6591 0.6628 0.6664 0.6700 0.6736 0.6772 0.6808 0.6844 0.6879

…

1.10 0.8643 0.8665 0.8686 0.8708 0.8729 0.8749 0.8770 0.8790 0.8810 0.883

1.20 0.8849 0.8869 0.8888 0.8907 0.8925 0.8944 0.8962 0.8980 0.8997 0.9015

1.30 0.9032 0.9049 0.9066 0.9082 0.9099 0.9115 0.9131 0.9147 0.9162 0.9177

1.40 0.9192 0.9207 0.9222 0.9236 0.9251 0.9265 0.9279 0.9292 0.9306 0.9319

…

1.80 0.9641 0.9649 0.9656 0.9664 0.9671 0.9678 0.9686 0.9693 0.9699 0.9706

1.90 0.9713 0.9719 0.9726 0.9732 0.9738 0.9744 0.9750 0.9756 0.9761 0.9767

2.00 0.9772 0.9778 0.9783 0.9788 0.9793 0.9798 0.9803 0.9808 0.9812 0.9817

2.10 0.9821 0.9826 0.9830 0.9834 0.9838 0.9842 0.9846 0.9850 0.9854 0.9857

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A variable is normally distributed with a mean of 2.00 and a variance of 16.00.

Using the excerpt, the probability of observing a value of 7.40 or less is closest

to:

A 、63.3%.

B 、91.2%.

C、 96.8%.

2、Which of the following most accurately describes how to standardize a random variable X?

A、 Subtract the mean of X from X, and then divide that result by the standard deviation of the standard normal distribution.

B 、Divide X by the difference between the standard deviation of X and the standard deviation of the standard normal distribution.

C 、Subtract the mean of X from X, and then divide that result by the standard deviation of X.

查看第1题答案解析

【答案】

【解析】

B is correct. First the outcome of interest, 7.40, is standardized for the given normal distribution:Z X = − ( ) µ σ = − ( ) 7 4. . 0 2 00 16 = 1 3. 5 Then, the given table of values is used to find the probability of a Z-value being less than or equal to 1.35 standard deviations above the mean. The value is P(Z ≤ 1.35) = 0.9115 = 91.2%.

A is incorrect; it divides 5.4 (that is the result of 7.4 – 2) by the variance, 16, and uses 0.34 as the z-value: P(Z≤0.34) = 0.6331 = 63.3%.

C is incorrect; it divides the value, 7.4, by the standard deviation, 4, and uses 1.85 as the Z-value: P(Z ≤ 1.85) = 0.9678 = 96.8%

查看第2题答案解析

【答案】

【解析】

C is correct. There are two steps in standardizing a random variable X: Subtract the mean of X from X, and then divide that result by the standard deviation of X. This is represented by the following formula: Z = (X – μ)/σ.

A is incorrect. There are two steps in standardizing a random variable X: Subtract the mean of X from X, and then divide that result by the standard deviation of X. This is represented by the following formula: Z = (X – μ)/σ.

B is incorrect. There are two steps in standardizing a random variable X: Subtract the mean of X from X, and then divide that result by the standard deviation of X. This is represented by the following formula: Z = (X – μ)/σ