Probability Distributions: Chapter 6 Statistics Homework Solution

This assignment provides solutions to homework problems related to Chapter 6, which focuses on continuous probability distributions. The problems cover topics such as uniform distribution, normal distribution, z-scores, and the central limit theorem. Detailed solutions are provided for each problem, including calculations and explanations. The homework explores concepts like finding probabilities within given ranges, determining z-values, and assessing whether certain events are unusual based on calculated probabilities. The document also includes an analysis of whether a given rainfall data is normally distributed, and examines the impact of sample size on the shape and standard deviation of sampling distributions. It concludes with a discussion on the implications of the central limit theorem and its application to various scenarios, such as blood pressure and dishwasher lifespan. Desklib is your go-to platform for more solved assignments and study resources.

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Section 6-1 Homework
2) a) The random variable is the waiting time during peak rush hour period for the commuter
train on the Red line at Cleveland. Height of the uniform distribution = (1/(8-0)) = 0.125
c) P(4<x<5) = (5-4)*0.125 = 0.125
d) P(3<x<8) = (8-3)*0.125 = 0.625
e) In the given case there would be only one line whose width would be zero, hence P (X=5)
= 0.*125 = 0
Section 6-3 Homework
2) a) Requisite z value = invNorm(0.15, 0,1) = -1.036
b) Area to the right is 65% which implies area of the left is (100-65) = 35%
Hence, requisite z value = invNorm(0.35, 0,1) = -0.385
c) Requisite z value = invNorm(0.1, 0,1) = -1.282
d) Area to the right is 5% which implies area of the left is (100-5) = 95%
Hence, requisite z value = invNorm(0.95, 0,1) = 1.645
e) The corresponding –z would be such that area to the left would be 2.5% of the total values.
Hence, requisite z value = invNorm (0.025, 0,1) = -1.96
Thus, the respective values are -1.96 and 1.96.
f) The corresponding –z would be such that area to the left would be 0.5% of the total values.
Hence, requisite z value = invNorm (0.005, 0,1) = -2.576
Thus, the respective values are -2.576 and 2.576.
4) a) The random variable is the blood pressure for people in China.

b) Sample mean = 128mm Hg, Standard deviation = 23
Z statistic = (135-128)/23 = 0.3043
Requisite probability = 1- P(Z<0.3043) = 1-0.6196 = 0.3804
c) Z statistic = (141-128)/23 = 0.5652
Requisite probability = P(Z<0.5652) = 0.7140
d) Z1 = (120-128)/23 = -0.3478
P (Z < Z1) = 0.3640
Z2 = (125-128)/23 = -0.1304
P (Z < Z2) = 0.4481
Hence, requisite probability = 0.4481 – 0.3640 = 0.0841
e) No, it would not be unusual to find a person in China with mean blood pressure exceeding
135mm Hg since the underlying probability (0.3804) is greater than 5% or 0.05.
f) Area to the left is 90%
Hence, requisite value = invNorm(0.9,128,23) = 157.5 mmHg
8) a) The random variable is the life of the dishwasher.
b) Mean life = 12 years, Standard deviation = 1.25 years
Z = (15-12)/1.25 = 2.4
P(Z>2.4) = 1- P(Z≤2.4) = 1-0.9918 = 0.0082
c) Z = (6-12)/1.25 = -4.8
P(Z<-4.8) = 0.0000
d) Z1 = (8-12)/1.25 = -3.2
P(Z<-3.2) = 0.0007

Z2= (10-12)/1.25 = -1.6
P(Z<-1.6) = 0.0548
Requisite probability = 0.0548 -0.0007 = 0.0541
e) If a dishwasher that lasted lower than 6 years is found, then clearly there is problem with
the manufacturing problem as it is almost impossible to find such a dishwasher considering
the probability of almost zero.
f) The warranty period should be such that 95% of the values should lie on the right and only
5% values on the right.
Requisite warranty period = invNorm(0.95,12,1.25) = 14.06 years
10) a) The random variable is yearly rainfall in Sydney.
b) Sample mean = 137 mm, Standard deviation = 69 mm
Z =(100-137)/69 = -0.5362
P (Z<-0.5362) = 0.2959
c) Z= (240-137)/69 = 1.493
P(Z>1.493) = 1-P(Z≤1.493) = 0.068
d) Z1 = (140-137)/69 = 0.0435
P(Z<0.0435) = 0.5173
Z2= (250-137)/69 = 1.638
P(Z<1.638) = 0.9493
Requisite probability = 0.9493 – 0.5173 = 0.432
e) Even if the rainfall is less than 100mm, it does not mean that it is an unusually dry year
considering the fact that associated probability with the happening of the same is greater than
0.05 and hence such an event is not rare or unusual.

f) Since the given amount should be such that 90% values have to be greater, then the area to
the left of the desired value would be 10% or 0.1.
Hence, requisite value = invNorm(0.1,137,69) = 48.57 mm or 49 mm (rounded off)
Section 6.4 Homework
4) The given rainfall data is not normally distributed for the following reasons.
 The histogram has a rightward skew or positive skew owing to which the shape is
asymmetric.
 The mean, median and mode do not coincide. Mean = 136.91, Median = 133.1 and
Mode= 90.9
 There is presence of outliers on the higher side which can be established from the
following computations.
One observation in the given rainfall data is 383 mm which exceeds the upper limit of 306.75
mm.
Section 6-5 Homework
2) a) If the sample size is 10, then the shape of the distribution would be symmetric about the
mean and bell shaped as when a sample is chosen from the normal population, the sample
distribution would be normal even for lower population size i.e. less than 30.
b) The sample mean would be the same as population mean i.e. 245. However, standard
deviation of sample = 21/√10 = 6.64
c) Z statistics = (241-245)/6.64 = -0.602
Requisite probability = 1- P(Z<-0.602) = 1-0.2735 = 0.7265

d) If the sample size is 35, then the shape of the distribution would be in the form of
symmetric bell shape curve since the given sample can be assumed to be normally distributed
in accordance with Central Limit Theorem.
e) The sample mean would be the same as population mean i.e. 245. However, standard
deviation of sample = 21/√35 = 3.55
f) The requisite computations are shown below.
Z statistics = (241-245)/3.55 = -1.13
Requisite probability = 1- P(Z<-1.13) = 1-0.1299 = 0.8701
g) The answer in part (d) is smaller than comparative answer in (f) which is on account of a
higher standard deviation in comparison to part (f) when the sample size has increased from,
10 to 35.
4) a) The random variable is the blood pressure for people in China.
b) The shape of distribution of the sample mean would be bell shaped and symmetric about
the mean since despite a lower size of the sample, it would be considered as a normal
distribution as per the Central Limit Theorem (CLT)
c) As per the CLT, sample mean would be same as population mean or 128mmHg.
d) As per the CLT, sample standard deviation = 23/√15 = 5.94 mmHg
e) Z statistic = (135-128)/5.94 = 1.179
Requisite probability = 1- P(Z<1.179) = 1-0.8807 = 0.1193
f) No, it would not be unusual to find a sample of 15 people in China with mean blood
pressure exceeding 135mm Hg since the underlying probability (0.1193) is greater than 5%
or 0.05.
g) It is apparent that the probability of obtaining such a sample even though not rare is quite
small. Hence, if such a sample is found, then it reflects at potential blood pressure related
issues with the Chinese people.

6) a) The random variable is the cholesterol levels of women age 45-59 in Ghana, Nigeria,
and Seychelles.
b) Mean cholesterol level = 5.1mmol/L, Standard deviation = 1mmol/L
Z = (6.2-5.1)/1 = 1.1
P(Z>1.1) = 1-P(Z≤1.1) = 0.1357
c) Here, the sample size = 2, hence sample standard deviation = 1/√2 = 0.707
Z = (6.2-5.1)/0.707 = 1.56
P(Z>1.56) = 1-P(Z≤1.56) = 0.0599
d) Here, the sample size = 3, hence sample standard deviation = 1/√3 = 0.5774
Z = (6.2-5.1)/0.5774 = 1.91
P(Z>1.91) = 1-P(Z≤1.91) = 0.0284
e) Considering that the probability of having a cholesterol level in excess of 6.2 mmol/L after
three tests is quite rare (probability less than 0.05) and thus it can be derived that cholesterol
level of women exceeds that of normal people.
8) a) The random variable is the life of the dishwasher.
b) As per the Central Limit Theorem, the sample mean would be equal to the population
mean of 12 years.
c) As per the Central Limit Theorem, the standard deviation of the sample mean = 1.25/√10 =
0.3953
d) The shape of sampling distribution is symmetric, bell shaped distributed about the mean
which is characteristic of normal distribution. This is because even though the sample size is
small but considering that the data has been selected from a normal population, hence the
sample would also be normally distributed.

e) Z = (6-12)/0.3953 = -15.18
P(Z<-15.18) = 0.0000
f) It is apparent from the above computation that the probability of finding a sample of 10
dishwashers with mean life less than 6 years is almost zero and hence this event is very rare.
Thus, the presence of such a sample would indicate some serious defect with the
manufacturing process or else such a sample should not exist.