Statistics Homework: Statistical Analysis, Sampling & Distributions

This statistics assignment provides solutions to problems involving the Poisson distribution, sampling techniques, and data analysis. The first question analyzes the Poisson distribution, including a probability table and graph. The second question explores population standard deviation and probabilities related to sample averages. The third question details how to generate a random sample of fourth-year students using Excel, as well as the problems associated with random samples. The fourth question focuses on measures of central tendency, frequency distribution, and histogram interpretation, including identifying the most frequent class interval and describing the distribution's skewness. The assignment demonstrates the application of statistical concepts and techniques to solve real-world problems.

Statistics 1
Name:
Institution:
Date:

Statistics 2
QUESTION ONE
a. Poisson distribution model
b. 0.003698
c. 0.97
d. 1- (0.024) = 0.976
e. Table and graph of poisson distribution
No. of complaints
Probabilit
y
0
0.003697
9
1
0.024405
9
2
0.082388
4
3
0.190622
4
4 0.34215
5
0.511860
9
6
0.670257
8
7
0.796975
3
8
0.885677
5
9 0.94087
10
0.971777
8
11
0.987512
7
12
0.994855
7
13
0.998018
8
14 0.999284
15
0.999756
4
16
0.999921
7
17
0.999976
2
18
0.999993
1
19
0.999998
1
20
0.999999
5
21
0.999999
9
22 1
23 1

Statistics 3
24 1
25 1
Table 1
1 2 3 4 5 6 7 8 9 10111213141516 17181920212223242526
0
5
10
15
20
25
30
0
0.2
0.4
0.6
0.8
1
1.2
No. of complaints
Probability
Figure 1
QUESTION TWO
Mean=15
δ =5
n=100
a) Population standard deviation;
standard deviation= √ variance= √ 25=5
b) Probability that sample average is more than 25
P( x >25)

Statistics 4
The formula is,
Z= x −μ
σ
μ =15
σ =¿25
p ( x>25 )=z= 25−15
5 =2
Z=¿2, This Z score corresponds to 97.72%
Probability that sample average is more than 25 = (100-97.72) %
= 2.28%
c) Probability that the sample is exactly 15
P( x=15)
The formula is,
Z= x −μ
σ
μ =15
σ =¿5
p ( x=15 ) =z =15−15
5 =0
Z=¿0, This Z score corresponds to 50%
Probability that sample average is exactly 15 = 50%
d) Population standard deviation?
Mean=15
δ=?
n=100
probability that mean is13=8 %

Statistics 5
The Z-score that corresponds to 8% from the normal table is 0.2
Therefore δ= x−μ
Z
δ = 13−15
0.2
=10
e) Population mean
Population Mean=?
δ =5
n=100
probability that sample meanis less than 12=10 %
Making population mean the subject of the formula, we have;
μ= x−Zδ
The value of Z that corresponds to 10% from the normal table is 0.26.
So, μ=12−(0.26× 5)
μ=12−1.3
μ=10.7
QUESTION THREE
A. How to pick a sample from fourth year students in excel.
 Have a list of all fourth year students in column B
 To the right of column B i.e C1 enter student number
 Click in the cell below C1 i.e C2, press (ctrl, shift, end)
 Type the formula =RAND() then press ctrl enter button
 Select cell C1
 Then click the sort button on the toolbar (AZ sort)

Statistics 6
 By doing the above, column C will correspond a random number of between 0 and 1 for
every data point in column B.
 The worksheet is then sorted by column C which causes an automatic sorting of column
B.
 The sample can now be chosen accordingly.
B. Creating a random sample in excel.
 Click on let say A1
 On this cell write the function =RAND(x, y) where x and y are your minimum and
maximum values of your data respectively.
 Click enter
 Putting your cursor on A1, hold and drag it to let’s say A20 if you needed 20 data points
C. Problems associated to random samples
 It is a biased method as not all elements of the population has an equal chance of being
selected to the sample.
 There is always a problem in estimating the sampling variance when it comes to non-
random sampling.
QUESTION FOUR
a. Measures of central tendency
Measures of central tendency
Mean 60.75520833
Median 30
Mode 5
Table 2
b. Frequency distribution and histogram

Statistics 7
 Find the smallest and the largest value
 Decide how many bins are needed (9)
 Calculate the range of the data
 Divide the range by the number of bins
 Create the bin boundary by starting with the smallest number and adding the bin size.
The figure below is of child mortality histogram
3 to 34 35 to 66 67 to 98 99 to
130 131 to
162 163 to
194 195 to
226 227 to
258 259 to
290
0
20
40
60
80
100
120
Child mortality histogram
Series1
Figure 2
Frequency distribution table of child mortality

Statistics 8
3 to 34 35 to 66 67 to 98 99 to
130 131 to
162 163 to
194 195 to
226 227 to
258 259 to
290
0
20
40
60
80
100
120
Figure 3
c. The most occurring class interval is 3 to 34. The distribution is not normal. It is skewed
to the right. This can be observed from the histogram (figure 2).