homework3

Question 3.2

normal.plot <- ggplot(data=data.frame(x=c(-3,3)), aes(x)) + stat_function(fun=dnorm, args=list(mean=0, sd=1))
normal.plot + stat_function(fun=dnorm, xlim=c(-1.13, 3), geom='area')

pnorm(q=-1.13, mean=0, sd=1, lower.tail=FALSE)

## [1] 0.8707619

• 1. \(\approx\) 87%

normal.plot + stat_function(fun=dnorm, xlim=c(-3, .18), geom='area')

pnorm(q=.18, mean=0, sd=1)

## [1] 0.5714237

• 2. \(\approx\) 57%

normal.plot + stat_function(fun=dnorm, xlim=c(3, 3), geom='area')

pnorm(q=8, mean=0, sd=1, lower.tail=FALSE)

## [1] 6.220961e-16

• 3. \(\approx\) 0%

normal.plot + stat_function(fun=dnorm, xlim=c(-.5, .5), geom='area')

pnorm(q=.5, mean=0, sd=1) - pnorm(q=-.5, mean=0, sd=1)

## [1] 0.3829249

• 4. \(\approx\) 38%

Question 3.4

• 1. Men’s: \(N(4313, 583)\) Women’s: \(N(5261, 807)\)
• 2. Leo’s Z-score: \(\frac{4948-4313}{583} \approx 1.09\) Mary’s Z-score: \(\frac{5513-5261}{807} \approx 0.31\)

Leo’s finishing time z-score is 1.08 meaning that his time is about 1 standard deviation above the mean finishing times in his group. Mary’s is 0.31 meaning that she is approximately 0.31 standard deviation above her groups finishing time.

• 3. Mary performed better than Leo. While both times are bove average, her’s has a smaller z score indiciating that she performed better than a larger proportion of her group than Leo did of his group.

pnorm(q=4948, mean=4313, sd=583, lower.tail=FALSE)

## [1] 0.1380342

• 4. \(\approx\) 14%

pnorm(q=5513, mean=5261, sd=807, lower.tail=FALSE)

## [1] 0.3774186

• 5. \(\approx\) 37%
• 6. b would remain the same. A z-score is a z-score. The formula does not change based on the distribution. c through e however would change. The linking of z score to a certain proportion of the population is contingent on the data being roughly normally distributed. Without this information there is no consistant way to use the z-score to determine the porportion of racers Leo or Mary performed better than.

Question 3.18

• 1. Yes, the data adhears to the 68-95-99.7 rule

female.heights <- c(54, 55, 56, 56, 57, 58, 58, 59, 60, 60, 60, 61, 61, 62, 62, 63, 63, 63, 64, 65, 65, 67, 67, 69, 73)
length(female.heights[female.heights > 56.94 & female.heights < 66.1]) / 25

## [1] 0.68

Z-score for 68% is -1 to 1 corresponding to a range of (56.94, 66.1)

length(female.heights[female.heights > 52.36 & female.heights < 70.69]) / 25 #96%

## [1] 0.96

Z-score for 95% is -1 to 1 corresponding to a range of (52.36, 70.69)

length(female.heights[female.heights > 47.78 & female.heights < 75.26]) / 25 #68%

## [1] 1

Z-score for 99.7% is -1 to 1 corresponding to a range of (47.78, 75.26)

qqnormsim(female.heights)

• 2. The actual data looks very similar to the simulated data. Most of the values fit nicely on the normal probability plot line with a few upper and lower outliers. It is safe to say that this data is roughly normal.

Question 3.22

dgeom(9, .02)

## [1] 0.01667496

• 1. 9 failures followed by a success at 2% has \(\approx\) 1.6% chance of occuring

pgeom(100, .02)

## [1] 0.8700328

• 2. \(\approx\) 87% chance of 100 successes in a row
• 3. Mean = \(\frac{1}{.02}\) =50 SD = \(\frac{\sqrt{1-.02}}{.02} \approx\) 49.49
• 4. Mean = \(\frac{1}{.05}\) =20 SD = \(\frac{\sqrt{1-.05}}{.05} \approx\) 19.49
• 5. As the probability of success (finding a defective piece) decreases, the mean increases. This makes sense as a lower chance of success would indicate more trials before a success. The standard deviation follows this logic as well.

Question 3.38

dbinom(2, 3, .51)

## [1] 0.382347

• 1. \(\approx\) 38%
• 2. MMF, MFM, FMM 0.51 \(\times\) 0.51 \(\times\) 0.49 + 0.51 \(\times\) 0.49 \(\times\) 0.51 + 0.49 \(\times\) 0.51 \(\times\) 0.51 \(\approx\) 38%
• 3. There is no need to perform seperate calculations for each scenario, account for disjoing probabilities, and sum those values. This would be tedious, repetitive work.

Question 3.42

dnbinom(7, 3, .15)

## [1] 0.03895012

• 1. \(\approx\) 3.9%
• 2. 15%. The trials are independent
• 3. The probability calculated in part b is not equivalent to that in part a. We cannot ignore the fact that her 10th serve will, in some scenarios, have a 0% chance of being her 3rd successful serve.