Probability and Statistics: Solving Problems on IQ, Colds, and More

This document provides solutions to a series of statistics and probability problems. It includes calculations and interpretations related to IQ scores, using normal distribution to find proportions above or below certain scores, and within certain ranges of the mean. Further problems cover the duration of common colds, calculating probabilities related to their length, and determining the number of hours suffered by specific percentiles. The solutions also address probability in other contexts, such as the performance of nuclear reactors, extrasensory perception (ESP) experiments, and the likelihood of breast cancer given mammogram results. Desklib is a valuable platform to find these solved assignments and study resources for students.

Chapter 58
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5.11. Question: IQ score
Mean(μ) =100, standard deviation(σ ) =15
a. Proportion above Kristen’s 125
P=1−P (z < 125−100
15 )=1−P ( z< 1.67 )
¿ 1−0.9525
¿ 0.0475
¿ 4.75 %
b. Below 82
P=P ( z< 82−100
15 )=P ( z ←1.2 )
¿ 0.1151
¿ 11.51 %
c. Within 9 points of the mean
This implies 9 points below the mean and 9 points above the mean: 100-9= 91 and
100+9=109
P=P ( z< 109−100
15 )−P ( z< 91−100
15 )
¿ P ( z< 0.6 )−P ( z ←.6 )
¿ 0.7257−0.2743
¿ 0.4515
¿ 45.15 %
d. More than 40 points from the mean

This implies 100+40=140
P=1−P (z < 140−100
15 )=1−P ( z<2.67 )=¿
1−0.9962=0.0038
¿ 0.38 %
5.13 Question
a. Upper 2 %=98%
x p=zσ +μ
x.98=2.054 ( 15 ) + 100
¿ 130.81
b. 10% lower
x.10=−1.2816 ( 15 ) +100
¿ 80.78
c. 60% upper
x.40=−0.25 ( 15 )+100=96.25
d. Middle 95%
x.95 upper=1.96 ( 15 ) +100=129.4
x.95lower=−1.96 (15 )+ 100=70.6
Therefore middle 95% will be bordered by
70.6 ≤ x ≤129.4
e. Middle 99%

x.99 upper=2.575 ( 15 ) +100=138.63
x.99 lower=−2.575 ( 15 )+100=61.38
Therefore middle 99% will be bordered by,
61.38 ≤ x ≤138.63
5.12. Question
Mean (μ) =83 hours, standard deviation (σ ) =20 hours
a. Estimated number of hours for the shortest-suffering 5%
In this case the score will be computed
x0.05=zσ + μ
¿−1.645 ( 20 )+ 83
¿ 50.103 hours
b. Proportion of sufferers whose colds latest longer than 48 hours
P=1−P (z < 48−83
20 )
¿ 1−P ( z ←1.75 )=1−0.0401
¿ 0.9599
¿ 95.99 %
c. Proportion that suffered few than 61 hours
P=P ( z< 61−83
20 )=P ( z ←1.1 )=0.1357

d. Number of hours suffered , 1% either above or below the mean
x0.01 above=zσ =2.3263 ( 20 ) =46.53
μ− zσ ≤ X ≤ μ+ zσ
¿ 83−46.53 ≤ X ≤83+46.83
36.47 ≤ X ≤129.53
e. Proportion that suffered between 24 hours(1 day) and 72 hours(3 days)
P=P ( z< 72−83
20 )−P ( z< 24−83
20 )
¿ P ( z←0.55 ) −P ( z←2.95 )
¿ 0.2912−0.0016
¿ 0.2896=28.96 %
f. Number of hours suffered by the middle 95%
X 0.95upper =1.96(20)+83=122.20
X 0.95lower =−1.96 ( 20 ) +83=43.80
Therefore, the middle 95% will be bordered by
43.80 hours ≤ x ≤122.20 hours
g. Proportion that suffered between 2( 48 hours) and 4 days(96 hours)

P=P ( z< 96−83
20 )−P (z < 48−83
20 )
¿ P ( z< 0.65 )−P ( z ←1.75 )=0.7422−0.0401
¿ 0.7021=70.21 %
h. Hours suffered by 20%
X 0.20=zσ +μ=−0.842 ( 20 ) +83
¿ 66.17 hours
i. Another researcher with 3% extreme, interested with those who suffered least and most
−1.88 ( 20 ) +83<x >1.88 ( 20 ) +83
45.38 ≤ x ≤ 120.62
Mild suffered for fewer than 45.38% and severe suffered for more than 120.62 hours
j. Proportion that will suffer for more than 61 hours
P=1−P ( 61−83
20 )=1−P ( z ←1.1 )
¿ 1−0.1357=0.8643
¿ 86.43 %
k. The portion that suffered for exactly 61 hours
The proportion at this point is 0 as the distribution is continuous (normal) which only
consider probabilities as area under a curve between two points

5.18 Question
Mean (μ) =28, standard deviation (σ ) =4
a. Mean is 50%
X 0.5=zσ +μ=0 ( 4 ) +28
¿ 28
Therefore median is equal to the mean
b. Z score that define overweight
( 25−28
4 ) ≤ Z ≤ ( 29.9−28
4 )
−0.75 ≤ z ≤ 0.475
c. Z score that defines obese
z > 30−28
4
z >0.5
8.10 Question
a. whether random or not
The poll was not random
b. improvement
By allowing people to call any number and do vote randomly
8.14 Question
Probability that a boy is born 0.5 also for a girl to be born is 0.5
a. P(BB)
P=0.5∗0.5=0.25

b. P(GG)
P=0.5∗0.5=0.25
c. Either two boys or two girls
P=P ( BB )+ P ( ¿ ) =0.25+ 0.25=0.5
8.16. Question
Extrasensory perception (ESP) possibilities – circle, Square, Cross, star, Wavy line
Probabilities:
a. C
P ( C ) = Number of correct outcome
Number of possible outcomes =1/5
From this the P ( I ) =1− 1
5 = 4
5
b. CI
P ( CI )=
1
5∗4
5 = 4
25
c. CCC for three guesses
P ( CCC )=( 1
5 )3
= 1
125
d. III for three guesses
P ( III ) = ( 4
5 )
3
= 64
125

8.19 monitoring the performance of nuclear reactors
False alarm or missed alarm
a. probability that sensor is incorrect
P=P ( false alarm )+P ( missed alarm )=0.02+0.01=0.03
b. Probability of both false alarm and missed alarm
P=P ( false alarm )∗P ( missed alarm ) =0.02∗0.01=0.0002
c. Probability of excess radiations
P ( Excess radiations )=0.98∗0.99=0.9702
18.21 Question
a. Probability of mammogram
P ( m|c )+P ( m|c' ) = ( 0.01∗0.8 )∗( 0.99∗0.1)=0.00079
b. Probability of having breast cancer given positive mammogram
P=0.01∗0.8=0.008
c. Probability of not having breast cancer given negative mammogram,
P= ( 0.01∗0.2 ) +(0.99∗0.90)=0.893