"quantity"exercise:standardized random variable

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

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%.

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