Practice Problems 15

Problem 1: Student Survey

This example explores the survey dataset from the MASS package, with a focus on the Height and Age variables. First, let’s examine the survey dataset, paying special attention to Height and Age:

survey <- MASS::survey # load the data
summary(survey$Age)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  16.75   17.67   18.58   20.37   20.17   73.00 
summary(survey$Height)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.    NA's 
  150.0   165.0   171.0   172.4   180.0   200.0      28

(a). Proportion Below a Given Value

Q1: What proportion of individuals have a height below 160 cm, based on our sample?

Hint: Calculate the mean and standard deviation for the Height variable first.

Click for answer 
# Mean and standard deviation for Height
Height_mean <- mean(survey$Height, na.rm = TRUE)
Height_sd <- sd(survey$Height, na.rm = TRUE)

# Proportion below 160 cm
pnorm(160, mean = Height_mean, sd =Height_sd)
[1] 0.1043305


(b). Determining a Specific Percentile

Q2: What is the age cutoff for the lower 75% of our sample distribution?

Hint: Determine the mean and standard deviation of the Age variable to find the age at the 75th percentile.

Click for answer 
# Mean and standard deviation for Age
age_mean <- mean(survey$Age, na.rm = TRUE)
age_sd <- sd(survey$Age, na.rm = TRUE)

# Age at the 75th percentile
qnorm(0.75, mean = age_mean, sd = age_sd)
[1] 24.74139


(c). Calculating the Middle 95 Percentile for Age

Q3: What are the age cutoffs that define the middle 95% of our sample distribution?

Hint: Calculate the 5th and 95th percentiles for the Age variable to find the ages that bound the middle 95% of the distribution.

Click for answer 
# Age at the 25th percentile
age_25th <- qnorm(0.05, mean = age_mean, sd = age_sd)
age_25th
[1] 9.725181
# Age at the 75th percentile
age_75th <- qnorm(0.95, mean = age_mean, sd = age_sd)
age_75th
[1] 31.02385


Problem 2: SAT Verbal scores

Suppose that the verbal SAT scores in a population are normally distributed with a mean \(\mu=580\) and standard deviation \(\sigma = 70\). If \(X\) is shorthand for a verbal SAT score, then we can write this as \(X \sim N(580,70)\).

(a) What proportion of scores are above 650?

Click for answer

Answer: About 15.9% of the scores are above 650.

pnorm(650,mean=580,sd=70) # proportion below
[1] 0.8413447
1-pnorm(650,mean=580,sd=70) # proportion above
[1] 0.1586553


(b) What is the 25th percentile (Q1)?

Click for answer

Answer: The score of about 533 is the 25th percentile, meaning 25% of the scores are below this value.

qnorm(.25,mean=580,sd=70)
[1] 532.7857


(c) What is the IQR for verbal SAT scores in this population? (Hint: find Q1 and Q3)

Click for answer

Answer: The 25th percentile (Q1) is 533 and the 75th percentile (Q3) is 627. The IQR for this normally distributed variable is about 94 points.

q1 <- qnorm(.25,mean=580,sd=70);q1
[1] 532.7857
q3 <- qnorm(.75,mean=580,sd=70);q3
[1] 627.2143
q3-q1
[1] 94.42857


(d) What score, high or low, will be deemed an outlier according the boxplot rules for outliers?

Click for answer

Answer: Using the 1.5IQR’s boxplot rule gives a lower fence of 392 and an upper fence of 768. So any score below 392 and above 768 will be called an outlier according to this rule.

1.5*94
[1] 141
q1 - 1.5*94
[1] 391.7857
q3 + 1.5*94
[1] 768.2143


(e) What percent of the population will be deemed an outlier?

Click for answer

Answer: We need to find the proportion of scores below 392 and above 768. With this symmetric distribution, we find about 0.004 in both tails. About 0.8% of the population will be deemed outliers according to the boxplot rule.

pnorm(392,mean=580,sd=70)
[1] 0.003618747
1-pnorm(768,mean=580,sd=70)
[1] 0.003618747


Problem 3: Standard Normal

The standard normal distribution has a mean of 0 and standard deviation of 1.

(a) What percent of SAT scores are at least 1 standard deviation above average?

Click for answer 
pnorm(1)  # proportion below
[1] 0.8413447
1-pnorm(1) # proportion above
[1] 0.1586553
Answer: About 16% of scores will be at least 1 standard deviation above average. (Note that the score of 580+70 = 650 is 1 standard deviation above average.)


(b) How many standard deviations away from average is the 25th percentile of SAT scores?

Click for answer

Answer: The 25th percentile of SAT scores (or any normally distributed values) is 0.67 standard deviations below average. We could also find this value using our answer to (1b):

qnorm(.25)
[1] -0.6744898

\[ z = \dfrac{533 - 580}{70} = -0.67 \]

(533 - 580)/70
[1] -0.6714286