library(DATA606)
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## Please visit openintro.org for free statistics materials
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## Welcome to CUNY DATA606 Statistics and Probability for Data Analytics 
## This package is designed to support this course. The text book used 
## is OpenIntro Statistics, 3rd Edition. You can read this by typing 
## vignette('os3') or visit www.OpenIntro.org. 
##  
## The getLabs() function will return a list of the labs available. 
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## The demo(package='DATA606') will list the demos that are available.
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3.2 Area under the curve

  1. Z > −1.13 (b) Z < 0.18 (c) Z > 8 (d) |Z| < 0.5
# p(z > -1.13)
normalPlot(bounds = c(-1.13,4))

# p(z < 0.18)
normalPlot(bounds = c(-4,0.18))

# p(z > 8)

normalPlot(bounds = c(8,10))

# |Z| < 0.5

normalPlot(bounds = c(-.5,.5))

3.4 Triathlon times

  1. N(\(\mu\) = 4313 , \(\sigma\) = 583 )
#b)
Leo.z.score = (4948 - 4313) / 583.0 
Leo.z.score
## [1] 1.089194
Mary.z.score = (5513 - 5261 ) / 807 
Mary.z.score
## [1] 0.3122677
#c)
#Mary is better ranked

#mary 
round(pnorm(5513, 5261, 807), 4)
## [1] 0.6226
#leo
round(pnorm(4948, 4313, 583), 4)
## [1] 0.862

3.18 Heights of female college students

height = 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 ) 
#a)
# 68-95-99.7% Rule
pnorm(mean(height)+sd(height), mean = 61.52, sd = 4.58) - pnorm(mean(height)-sd(height), mean = 61.52, sd = 4.58)
## [1] 0.6830768
pnorm(mean(height) + sd(height) * 2, mean = 61.52, sd = 4.58) - pnorm(mean(height)-sd(height) *2, mean = 61.52, sd = 4.58)
## [1] 0.9546724
pnorm(mean(height)+sd(height) *3, mean = 61.52, sd = 4.58) - pnorm(mean(height)-sd(height) *3, mean = 61.52, sd = 4.58)
## [1] 0.9973214
#b)

qqnorm(height)
qqline(height)

qqnormsim(height)

#looking a the graph it seem nearly normal.

3.22 Defective rate

#a)
(0.98) ^ 9 * 0.02
## [1] 0.01667496
#b)

.98 ^ 100
## [1] 0.1326196
#c)

1/.02
## [1] 50
#d)

sqrt((1-.05)/(.05)^2)
## [1] 19.49359
#increasing the probability of an event decreased the waiting time.

3.38 Male children

#a)
n <- 3
x <-2
dbinom(x, n, 0.51)
## [1] 0.382347
#b)
#part a and b the same match
boy.p <- 0.51 
girl.p <- 0.49

(boy.p *boy.p * girl.p) + (boy.p * girl.p * boy.p) + (girl.p * boy.p * boy.p)
## [1] 0.382347
  1. Part b will be more tedious because of the amount of combination we have to do in order to to find all the different 8 probabilities

3.42 Serving in volleyball

#a)
p <- 0.15
k <- 3
n <- 10
round(choose(n - 1, k - 1) * ((1 - p)^ (n - k) * p^k),4)
## [1] 0.039

  1. Independent event and will be 15 %.

  2. That’s even the same scenario on part a we still count all 3 serves while on part we already have specific 2 successful serve in nine attempts which as independent event we just focus on the last success serve event.