Chapter 3

3.2)
a) Z > -1.13
12.92 percentile corresponds to Z = -1.13
100-12.92 = 87.08 %

b) Z < 0.18
57.14%

c) Z > 8
very, very low
less than 1 in a billion

d) |Z| < 0.5
-0.5 < Z < 0.5
30.58 percentile for Z = -0.5
69.15 percentile for Z = 0.5
38.57 %

3.4)
a) men N(\(\mu\) = 4313, \(\sigma\) = 583)
women N(\(\mu\) = 5261, \(\sigma\) = 807)

b) Leo  

Z_Leo <- (4948-4313)/583
Z_Leo
## [1] 1.089194

Mary 

Z_Mary <- (5513-5261)/807
Z_Mary
## [1] 0.3122677


Leo’s time was 1.09 standard deviations above the mean.
Mary’s time was 0.31 standard deviations above the mean.
c) Leo ranked higher in his respective group because his time was more standard deviations above the mean.
d) Leo finished faster than 86.21% of his group.
e) Mary finished faster than 62.17% of her group.
f) The Z scores would stay the same because Z scores can be calculated for distributions that are not normal. However the percentiles would change if the distribution was not normal so the answers to parts d and e would be different.

3.18)
a) 68% of data should fall between z = -1 and z = 1
z = (x - \(\mu\))/\(\sigma\)

x <- (-1*4.58) +61.52
x
## [1] 56.94
y <- (1*4.58) +61.52
y
## [1] 66.1


percentage of trials in that range = 17/25 = 68%
95% of data should fall between z = -2 and z = 2
z = (x - \(\mu\))/\(\sigma\)

z <- (-2*4.58) +61.52
z
## [1] 52.36
w <- (2*4.58) +61.52
w
## [1] 70.68

percentage of trials in that range = 24/25 = 96%
99.7% of data should fall between z = -3 and z = 3
z = (x - \(\mu\))/\(\sigma\)

v <- (-3*4.58) +61.52
v
## [1] 47.78
u <- (3*4.58) +61.52
u
## [1] 75.26

percentage of trials in that range = 25/25 = 100%
The data conforms closely to the 68-95-99.7 rule
b) The data does not quite follow a normal distribution. There is a lot of data concentrated around the mean and then it drops off dramatically. It is unimodal but the range is very wide. The data mostly falls on the normal probability plot but there is an outlier on the low and the high end. Based on the mismatch between the histogram and the normal plot, I would say that the data does not quite follow a normal distribution.

3.22)
a) P(10th defective) = (.98)^9 * (.02)^1
0.017

b) n = 100 k = 0
P(0 successes) = 100!/(0! * 100!) * (.02)^0 *(.98)^100
.13
c) # trials before 1 success
\(\mu\) = 1/p
\(\mu\) = 1/0.02 \(\mu\) = 50  

\(\sigma\) = sqrt((1-p)/p^2)
\(\sigma\) = sqrt((1-.02)/.02^2)
\(\sigma\) = 49.4974747
d) P(defective) = 0.05
\(\mu\) = 1/.05
\(\mu\) = 20
\(\sigma\) = ((1-0.05)/.05^2)
\(\sigma\) = 19.4935887

e) Increasing the probability of an event, decreases the mean and decreases the standard deviation.

3.38) a) P(boy) = 0.51 n = 3
k = 2
Pb = 3!/(2!1!) (.51)^2 * (1-.51)^1
0.382347
b) BBG, BGB, GBB
Pd = .51 x.51 x .49 + .51 x .49 x .51 + .49 x 51 x .51
0.382347
The probability values are the same.
c) n = 8 k = 3
Using the addition rule for disjoint outcomes would be tedious because there are 8 choose 3 outcomes.
8!/(3!5!) = 56 possible scenarios. That is a lot to draw.

3.42) P(making a serve) = 0.15
a) P(making a 3rd serve on the 10th try)
n = 10 k = 3
P = 9 choose 2 .15^3 * .85^7
P = 0.039
b) 0.15
c) In part b, the question is just asking for the probability of having a successful serve. Since each serve is independent of the previous, that is just the probability of making a serve. In part a, it is asking for the probability of having a particular number of successes in a particular number of trials with the last trial being a success. They are different questions.