Statistics Assignment - Module Name, Semester 1, University Name

This statistics assignment provides detailed solutions to various problems related to statistical concepts. The first answer explores probability calculations using the normal distribution, including the probability of a can containing less than a certain amount of liquid, and the probability for a sample of cans. The second answer calculates the probability of a sample requiring a service call within a year, based on a given proportion. The third answer focuses on calculating the mean and standard deviation of newspaper sales, and finding a 90% confidence interval for the mean. The fourth answer involves calculating confidence intervals for the average selling price of a house and determining the required sample size, along with the impact of confidence levels. Finally, the fifth answer calculates the required sample size for a survey and determines a confidence interval for the proportion of BMW owners within a specific age group. The assignment covers concepts such as confidence intervals, hypothesis testing, normal distribution, and sample size determination, with detailed calculations and explanations provided for each problem.

Running head: STATISTICS
Statistics
Name of the Student
Name of the University
Author note

1STATISTICS
Table of Contents
Answer 1..........................................................................................................................................2
Part a............................................................................................................................................2
Part b............................................................................................................................................2
Part c............................................................................................................................................3
Answer 2..........................................................................................................................................3
Answer 3..........................................................................................................................................3
Answer 4..........................................................................................................................................4
Part a............................................................................................................................................4
Part b............................................................................................................................................4
Part c............................................................................................................................................5
Answer 5..........................................................................................................................................6
Part a............................................................................................................................................6
Part b............................................................................................................................................6

2STATISTICS
Answer 1
Part a
The amount of Classy Cola in the cans follows a normal distribution
with X N ( μ=355.4 , σ =1.5)
Thus, the probability that a randomly selected can would have less than 355ml
P ( X <355 ) =P ( X−μ
σ < 355−355.4
1.5 )
¿ P(Z<−0.2667)
= 0.3949
Thus 39.49% of the cans have less than 355 ml of Classy Cola in its cans
Part b
The number of randomly selected cans = 50
Thus, the probability that 50 randomly selected can would have less than 355ml
P ( X <355 ) =P ( X−μ
σ / √ n < 355−355.4
1.5/ √ 50 )
¿ P(Z<−1.8856)
= 0.0297
Thus the probability that all 50 cans would have less than 355 ml of Classy Cola is 2.97%

3STATISTICS
Part c
The calculation of the probability for the number of cans in “part a” and “part b” was
based on the fact that the filling of the cans is normally distributed. If they do not follow normal
distribution then the results would not be valid. There are other types of distribution like
Binomial and Poisson distribution. However the use of these distributions are not valid when
filling of cans is considered.
Answer 2
The number of households surveyed = 4000.
The proportion of products that require service call in the first year = 0.03
Thus, P ( ^P>0.32 )=P
( ^P− p
√ p∗( 1− p )
n )=P
( 0.32−0.30
√ 0.30∗( 1−0.30 )
4000 )=P ( 0.02
0.007 )=P (2.76)
Hence, P ( ^P>0.32 ) =1−0.997=0.003
Thus, the probability that more than 32% of the sample require a service call within the first year
is 0.003 or 0.3%.
Answer 3
The total number of newspaper sold on the 10 days = 135+ 274 +209+ 399+ 153+ 321+
261 +414 +362 +269 = 2797
Thus the average number of newspapers sold on the 10 weekdays = x= 2797
10 =279.7
The standard deviation of the number of newspapers sold on the weekdays =
 = 96.17

4STATISTICS
Hence, the 90% confidence interval of the number of newspapers sold = x ± z∗σ
√ n
¿ 279.7 ± 1.65∗96.17
√10 =279.7 ± 1.65∗30.41=279.7 ±50.18
Thus, the 90% confidence interval of the mean number of newspapers sold on weekdays
= 229.52, 329.88 ≈ 230, 330
The conditions \under which the confidence interval is valid are :
1. The data has been taken from the newspaper stand owned by Noah Gnus at Ottawa.
2. The data pertains to the sales of the Newspaper “Ottawa Citizen.”
3. The sales of the newspapers is for weekdays.
Answer 4
Part a
The average selling price of a house at Ottawa = $405,600
The standard deviation of the selling price of homes = $67,500
The number of homes sold = 64
Hence, the 95% confidence interval of the selling price of a house = x ± z∗σ
√n
¿ 405600 ± 1.96∗67500
√64 =405600 ±1.96∗8437.5=405600 ±16537.5
Thus, the 95% confidence interval for the mean selling price of a house = 389062.5, 422137.5 ≈
$389, 063, $422, 138.
The confidence interval is valid for Ottawa.

5STATISTICS
Part b
The margin of error of the average selling price of a house = $10,000
Thus, z∗σ
√n = z∗67500
√n = 1.96∗67500
√n = 132300
√n = 10000
Thus, √n=1 3.23
n=175.03 ≈175
Thus the selling price of the number of houses required = 175
Part c
The z-value for 99% confidence interval is 2.576 whereas the z-value for 95% confidence is
1.96.
Hence in “part a” when using a 99% confidence interval the interval for the average selling price
of a house would become more wide. The lower limit would decrease in 99% confidence interval
as compared to 95% confidence interval. The upper limit would increase in 99% confidence
interval as compared to 95% confidence interval.
In “part b” the number of houses for 99% confidence interval would be more than 95%
confidence interval, since the z-value is more for 99% confidence interval as compared to 95%
confidence interval. The number of houses (n) required is calculated using the formula
n=( z−value∗σ
Margin of Error )
2
Thus keeping
 and “Margin of Error” constant the number of houses (n) increases with z-value.
Since the z-value for 99% confidence interval (2.576), is more than the z-value for 95%

6STATISTICS
Confidence interval (1.96) hence number of houses for 99% confidence interval would be more
than 95% confidence interval.
Answer 5
Part a
The margin of Error = 2% = 0.02
The z-value for 97% confidence interval = 2.17
Since, no estimate is given for the number of BMW owners between the age’s of 28 and 35
Hence, Let’s estimate that the proportion of BMW owners between the age’s of 28 and 35 years
be 0.5
Therefore the proportion of BMW owners not in the age group = 0.5
Thus the estimate is given by
0.02=2.17∗ √ pq
n =2.17∗
√ 0.5∗0.5
n
√n= 2.17∗0.5
0.02 = 0.1085
0.02 =5.425
n=29.43 ≈30
Hence, the number of BMW owners to be sampled = 30
Part b
The number of BMW owners surveyed = 120
The number of owners who are in the age group = 36

7STATISTICS
Thus the proportion of owners in the age group = 36
120 =0.3
Thus the proportion of owners who are not in the age group = 0.7
Thus, the 97% confidence interval = p ± z−value∗
√ pq
n =0.3 ± 2.17∗ √ 0.3∗0.7
120
¿ 0.3 ± 2.17∗0.042=0.3 ± 0.09
Thus, the proportion of BMW owners in the age group ranges between 0.21 and 0.39