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问题如下：

Question 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:

选项：

A.63.3%. B.91.2%. C.96.8%.

解释：

Solution

B is correct. First the outcome of interest, 7.40, is standardized for the given normal distribution:

Z=(X-μ/σ)=(7.40-2.00)/4=1.35?√=1.35$Z=\left(X?\mu /\sigma \right)=\left(7.40?2.00\right)/\sqrt{16}=1.35$

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%.