The Problem

Sophia who took the GRE scored 160 on the Verbal Reasoning section and 157 on the Quantitative Reasoning section. The mean score for Verbal was 151 with a standard deviation of 7, and the mean score for Quantitative was 153 with a standard deviation of 7.67. Suppose both distributions are nearly normal.

Answers in R code

######(a) Write down the short-hand for these two normal distributions

# Normal(151, 7) for Verbal Reasoning
# Normal(153, 7.67) for Quantitative Reasoning



######(b) What is her z-score on each? Draw it on a graph.

verbal_z <- (160 - 151) / 7
quant_z <- (157 - 153) / 7.67

verbal_z
## [1] 1.285714
quant_z
## [1] 0.5215124
curve(dnorm(x,0,1), xlim=c(-3,3), main="Z-Scores", xlab="z", ylab="Density")
abline(v=verbal_z, col="red")
abline(v=quant_z, col="blue")
legend("topleft", legend=c("Verbal", "Quantitative"), lty=c(1,1), col=c("red", "blue"))

######(c) Interpretation

# The z-scores tell us how many standard deviations her scores were higher or lower than average. In this case both were higher than average.



######(d) On which test did she do better?

# Relative to others she did better on Verbal, as evidenced by the higher z-score.



######(e) Find her percentile scores.

verbal_pct <- pnorm(verbal_z, lower.tail = T)
quant_pct <- pnorm(quant_z, lower.tail = T)

verbal_pct
## [1] 0.9007286
quant_pct
## [1] 0.6989951
######(f) What percent of test takers did better than her?

1 - verbal_pct
## [1] 0.0992714
1 - quant_pct
## [1] 0.3010049
######(g) Why is her raw score comparison different than z-score?

# In this case it is not, but it could be if the means and SDs were different.



######(h) If the distributions of scores were not nearly normal, would the answers above change?

# Yes, z-score analysis is based on an assumption of normality.