library(dplyr)
library(knitr)
library(visualize)
## Warning: package 'visualize' was built under R version 3.5.3

(4.1) Area under the curve, Part I

What percent of a standard normal distribution N (µ= 0; σ = 1) is found in each region? Be sure to draw a graph.

A. Z < -1.35

pnorm(-1.35, mean = 0 , sd = 1, lower.tail = TRUE)
## [1] 0.08850799
visualize.norm(stat = -1.35, mu = 0, sd = 1, section = "lower")

The percent of the distribution is .088507099% or 8.85%.

B. Z > 1.48

pnorm(1.48, mean = 0 , sd = 1, lower.tail = FALSE)
## [1] 0.06943662
visualize.norm(stat = 1.48, mu = 0, sd = 1, section = "upper")

The percent of the distribution is 0.06943662 or 6.94%.

C. -.44 < Z < 1.5 

pnorm(-.44, mean = 0 , sd = 1, lower.tail = TRUE) - 
pnorm(1.5, mean = 0, sd = 1, lower.tail = TRUE)
## [1] -0.6032242
visualize.norm(stat = c(-.44, 1.5), mu = 0, sd = 1, section = "bounded")

The percent of the distribution is .6032242 or 60.3%.

(4.3) GRE scores, Part I

Sophia who took the Graduate Record Examination (GRE) scored 160 on the Verbal Reasoning section and 157 on the Quantitative Reasoning section. 

The mean score for Verbal Reasoning section for all test takers was 151 with a standard deviation of 7, and the mean score for the Quantitative Reasoning was 153 with a standard deviation of 7.67. 

Suppose that both distributions are nearly normal.

** Write down the short-hand for these two normal distributions.**

The normal distribution shorthand is N(mean , standard deviation)

  1. The verbal reasoning section normal distribution is N(151, 7).
  2. The quantitative reason section normal distribution is N(153, 7.67).

What is Sophia’s Z-score on the Verbal Reasoning section? On the Quantitative Reasoning section?

The z-score formula is \[ Z-score = \frac {x - µ} {σ} \]

The Z-score for Sophia’s verbal reasoning is 1.285714.

(160 - 151) / 7
## [1] 1.285714

The Z-score for Sophia’s quantitative reasoning is .5215124.

(157 - 153) / 7.67
## [1] 0.5215124

Draw a standard normal distribution curve and mark these two Z-scores.

curve(dnorm, from = -4, to = 4)
abline (v = 1.285714, col = "green")
abline (v = 0.5215124, col = "blue")
text(1.285714, 0.3, "Verb Z: 1.29",col="green") 
text(0.5215124, 0.1, "Quan Z: 0.52", col="blue")

What do these Z-scores tell you?

The Z-score describes how far above or below the observation is to the described mean.

In context, the z-scores of Sophia’s test scores indicates that in her exams, she scored 1.29 standard deviations above the average in verbal reasoning test and .53 standard deviations above the average in the quantitative reasoning test.

Relative to others, which section did she do better on?

Sophia scored better on the verbal reasoning test by 1.29 standard deviations better than the mean.

Find her percentile scores for the two exams.

pnorm(160, mean = 151, sd = 7, lower.tail = TRUE )
## [1] 0.9007286

The percent of the distribution is .9007286 or 90%. Which means Sophia ranked in the 90th percentile of the exam test scores.

pnorm(157, mean = 153, sd = 7.67, lower.tail = TRUE)
## [1] 0.6989951

The percent of the distribution is .6989951 or 70%. Which means Sophia ranked in the 70th percentile of the exam test scores.

What percent of the test takers did better than her on the Verbal Reasoning section? On the Quantitative Reasoning section?

Since Sophia ranked in the 90th percentile in the verbal reasoning test, out of all test takers for that exam, (100-90) 10% did better than her.

Similarly, since Sophia ranked in the 70th percentile in the quantitative reasoning exam, which means out of all the test takers, (100-70) 30% did better than her.

Explain why simply comparing raw scores from the two sections could lead to an incorrect conclusion as to which section a student did better on.

The two test are not comparable because they have different metrics and different results which would not provide much use. It would like comparing a banana to a potato, which you could compare some characteristics however it would be more informational to compare banana to other fruits and potatoes to other root vegetables.

If the distributions of the scores on these exams are not nearly normal, would your answers to parts (b)-(f) change? Explain your reasoning.

The Z-scores would not change as it can be calculated from abnormal results, but nearly everything else, in regards to parts b-f, would because of the shift in a normal distribution to abnormal distributions which would require different evaluations.

(4.5) GRE scores, Part II.

In Exercise 4.3 we saw two distributions for GRE scores: N(µ = 151; σ = 7) for the verbal part of the exam and N(µ = 153; σ = 7:67) for the quantitative part. Use this information to compute each of the following:

The score of a student who scored in the 80th percentile on the Quantitative Reasoning section.

#z = qnorm(.8) = .84
quanmean = 153
quansd = 7.67
z80 <- (.84 * quansd) + quanmean

show(z80)
## [1] 159.4428

A person in the 80th percentile would have a min score of 159.4428.

The score of a student who scored worse than 70% of the test takers in the Verbal Reasoning section.

#z = qnorm(.3) = -.5244005, 70% worse in other words is 30% or less. 
verbmean = 151
verbsd = 7
z30 <- (-.52*verbsd) + verbmean

show(z30)
## [1] 147.36

A score of a student who scored worse than 70% of test takers is a min of 147.36.

(4.7) LA weather, Part I

The average daily high temperature in June in LA is 77°F with a standard deviation of 5°F. Suppose that the temperatures in June closely follow a normal distribution.

What is the probability of observing an 83 F temperature or higher in LA during a randomly chosen day in June?

pnorm(83, mean = 77 , sd = 5, lower.tail = FALSE)
## [1] 0.1150697
visualize.norm(stat = 83, mu = 77, sd = 5, section = "upper")

The probability of observing 83 F weather or higher temperature in LA is .1150 or 11.5%. The Z-score of this distribution is 1.2.

How cool are the coldest 10% of the days (days with lowest average high temperature) during June in LA?

#z = qnorm(.1) = -1.281552
#using the z-score formula = (-1.28 * 5) + 77 = 70.6
pnorm(70.6, mean = 77, sd = 5, lower.tail = TRUE)
## [1] 0.1002726
visualize.norm(stat = 70.6, mu = 77, sd = 5, section = "lower")

The temperature of the coolest 10% days in LA is around 70.6 F and less. The z-score of this distribution is -1.28.