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 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.
# 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] 0.1043305Q2: 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.
# 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] 24.74139Q3: 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.
# 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] 31.02385Suppose 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 \(X \sim N(580,70)\).
Answer: About 15.9% of the scores are above 650.
pnorm(650,mean=580,sd=70) # proportion below[1] 0.84134471-pnorm(650,mean=580,sd=70) # proportion above[1] 0.1586553Answer: 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] 532.7857Answer: 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] 532.7857q3 <- qnorm(.75,mean=580,sd=70);q3[1] 627.2143q3-q1[1] 94.42857Answer: 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] 141q1 - 1.5*94[1] 391.7857q3 + 1.5*94[1] 768.2143Answer: 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] 0.0036187471-pnorm(768,mean=580,sd=70)[1] 0.003618747The standard normal distribution has a mean of 0 and standard deviation of 1.
pnorm(1) # proportion below[1] 0.84134471-pnorm(1) # proportion above[1] 0.1586553Answer: 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