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Frequency Distribution Assignment
Question 1: For the Academic Community of Higher Education improving student's academic performance is not an easy task. A case study was conducted using descriptive Statistics which is the discipline of quantitatively describing the main features of a collection of information. A data of 200 Students marks as population was collected and out of it 25 students marks are being analyzed for the case study by applying some measures that can be used to describe a data set.
Let us consider the marks obtained from a sample of 25students listed as follows:
70 80 86 46 56 66 76 86 90 70 50 45
94 65 55 60 90 80 70 71 72 62 64 76 70
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Solution:
(i) Construct a frequency distribution & relative frequency distribution for the above data with 5 classes. (1)
|
Classes
|
Frequency
|
Relative frequency = frequency/total
|
|
45-55
|
3
|
0.12
|
|
55-65
|
5
|
0.2
|
|
65-75
|
7
|
0.28
|
|
75-85
|
5
|
0.2
|
|
85-95
|
5
|
0.2
|
|
|
25
|
1
|
(ii) Construct a Histogram with appropriate scale for the data. (1)
The following is the histogram:
(iii) Calculate the mean for the data using the frequency distribution table. (1)
The mean is as follows:
|
Classes
|
Frequency
|
Relative frequency = frequency/total
|
midpoint = (lower limit+upper limit)/2
|
midpoint * frequency
|
|
45-55
|
3
|
0.12
|
50
|
150
|
|
55-65
|
5
|
0.2
|
60
|
300
|
|
65-75
|
7
|
0.28
|
70
|
490
|
|
75-85
|
5
|
0.2
|
80
|
400
|
|
85-95
|
5
|
0.2
|
90
|
450
|
|
|
25
|
1
|
|
1790
|
Mean = 1790/25 = 71.6
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Question 2: Three males with an X-linked genetic disorder have one child each. The random variable x is the number of children among the three who inherit the X-linked genetic disorder.
|
x
|
0
|
1
|
2
|
3
|
|
P(x)
|
0.15
|
0.20
|
0.30
|
0.35
|
Determine whether it is a probability distribution or not. If it is a probability distribution, find its mean and standard deviation.
Solution:
It is a probability distribution as the sum of p(x) = 0.15+0.20+0.30+0.35 = 1 and each probability is between 0 and 1.
Mean=∑xpx=(0*0.15)+(1*0.20)+(2*0.30)+(3*0.35)=0+0.20+0.60+1.05=1.85
Standard deviation=√((x-mean)^2*p(x) )=√1.1275=1.06
x p(x) x.p(x) (x-mean)^2 ((x-mean)^2).p(x)
|
x
|
p(x)
|
x.p(x)
|
(x-mean)^2
|
((x-mean)^2).p(x)
|
|
0
|
0.15
|
0
|
3.4225
|
0.513375
|
|
1
|
0.2
|
0.2
|
0.7225
|
0.1445
|
|
2
|
0.3
|
0.6
|
0.0225
|
0.00675
|
|
3
|
0.35
|
1.05
|
1.3225
|
0.462875
|
|
|
|
1.85
|
|
1.1275
|
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Question 3: One Market Research company determined that 13% of college students work part-time during the academic year. For a random sample of 5 students, what is the probability that at least 3 students work part-time?
Solution:
n=5
p=0.13
q=1-p=0.87
Probability at least 3 works part time=P(3)+P(4)+P(5)
Required probability=C35 0.133 0.872 +C45 0.134 0.871 + C55 0.1350.870=0.18
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