STAT 02290 | 2026 Spring

Practice: Position in Distribution

Standard Normal Distributions

Determine the output of the following R code.

> pnorm(1.234)+pnorm(-1.234)
Answer:

Solution. The area represented by pnorm(1.234) is given as shown:

The area represented by pnorm(-1.234) is given as shown:

By symmetry, pnorm(-1.234) is equal to the area to the right of $z=1.23$:

So together, pnorm(1.234)+pnorm(-1.234) gives the entire area (namely, $1$) underneath the entire standard normal distribution.

Which number below is closest to the output of the following R code?

> pnorm(1)-pnorm(-1)

Solution. pnorm(1)-pnorm(-1) records the area between $z=-1$ to $z=1$ under the standard normal distribution. By the 68-95-99.7 Rule, this area is $\approx0.68$:

Which number below is closest to the output of the R code pnorm(-2)?

Solution.

By the 68-95-99.7 Rule, the area under the standard normal distribution from $z=-2$ to $z=2$ is $\approx0.95$:

Therefore its complement has area $\approx1-0.95=0.05$, half of which is $0.025$:

Which number below is closest to the output of the R code pnorm(2)?

Solution.

By the 68-95-99.7 Rule, the area under the standard normal distribution from $z=-2$ to $z=2$ is $\approx0.95$:

Therefore its complement has area $\approx1-0.95=0.05$, half of which is $0.025$:

The complement of this gives the area from $z=-\infty$ to $z=2$, which is $\approx1-0.025=0.975$:

Boxplots, Quartiles, IQR, Outliers

(AP Statistics Practice Exam)

Janelle collected data on the amount of time in minutes each person in a large sample of customers spent in a local store. The data also included recording the gender of each customer. These data were used to generate the boxplots shown below. Which of the following statements is true?

Boxplot of male and boxplot of female time spent in minutes

Solution.

(A) is false: the range is the maximum data value (for male, $\approx90$) minus the minimum data value (for male, $\approx5$).

(B) is false: from the boxplot, the male dataset without the outliers is roughly symmetric so its mean would be approximately equal to the median of $20$. However, the presence of the two extreme outliers at 60 and 90 skews the mean of the overall male dataset to be significantly larger than the median of $20$.

(C) is false: the male Q3 is approximately 29, not 45.

(D) is false: for females, IQR = Q3 - Q1 ≈ 17 - 8 = 9, not 15.

(E) is true: the male dataset to the right of the median consists of 50% the dataset, and the male median time is greater than every point of the female dataset.