SL 4.9—Normal distribution and calculations

Let $X$ be the mass of an apple. Then $X\sim N(180,15^2)$, so $\mu =180$ and $\sigma =15$.

We need $P(X<150)$. Notice that $150=180-2(15)=\mu -2\sigma$, which is exactly 2 standard deviations below the mean.

Using normalcdf(-1E99, 150, 180, 15) on a GDC gives $\approx 0.0228$. This is consistent with the empirical rule: about 95% of values lie within $\mu \pm 2\sigma$, so about 5% lie outside, and by symmetry about 2.5% lie below $\mu -2\sigma$.

The probability that a randomly selected apple is rejected is approximately 0.0228.

We need $P(X<150)$ where $X\sim N(180,15^2)$. The value 150 is 2 standard deviations below the mean.

The option $\approx 0.9772$ would represent $P(X<210)$, i.e., the area to the left of $\mu +2\sigma$. Since we want the area in the lower tail, this is far too large and is eliminated.

The value $\approx 0.1587$ corresponds to $P(X<\mu -\sigma )=P(X<165)$, i.e., one standard deviation below the mean. Since 150 is two standard deviations below the mean, this probability is too large and is eliminated.

The value $\approx 0.05$ is not the exact result here; it would approximate the total area in both tails beyond $\mu \pm 2\sigma$ combined. For just the left tail at exactly $\mu -2\sigma$, the precise value is about 0.0228, not 0.05.

The correct answer is $P(X<150)\approx 0.0228$, consistent with the lower tail beyond 2 standard deviations below the mean.

Let $X$ be the waiting time. Then $X\sim N(45,12^2)$, with $\mu =45$ and $\sigma =12$.

Check: $\mu -2\sigma =45-24=21$ and $\mu +2\sigma =45+24=69$. So 21 and 69 are exactly 2 standard deviations from the mean.

By the 68–95–99.7 rule, $P(\mu -2\sigma <X<\mu +2\sigma )\approx 0.95$. Therefore $P(21<X<69)\approx 0.95$.

The correct description is approximately 0.95, since 21 and 69 are each 2 standard deviations from the mean.

We need to identify which empirical rule percentage applies to the interval $(21,69)$ for $X\sim N(45,12^2)$.

The 68% rule applies to $\mu \pm 1\sigma =(33,57)$. Since $21\ne 33$ and $69\ne 57$, this option is incorrect.

The 99.7% rule applies to $\mu \pm 3\sigma =(9,81)$. Since the bounds are 21 and 69, not 9 and 81, this option is incorrect.

A probability of 0.50 would only apply to half the distribution (e.g., $P(X<45)$). Covering an interval from 21 to 69 — a wide range on both sides of the mean — clearly captures far more than 50% of the data.

Since $\mu \pm 2\sigma =(21,69)$, the empirical rule gives $P(21<X<69)\approx 0.95$.