Question

 

Q1: 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 25 students 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

 

1. Construct a frequency distribution & relative frequency distribution for the above data with 5 classes.            (1)                                                                   
2. Construct a Histogram with appropriate scale for the data.   (1)                   
3. Calculate the mean for the data using the frequency distribution table.     (1)

Q2: 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.

Q3: 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?

Answer 1

Q1)

a) Looking at the data the minimum is 45 and maximum is 94. Hence the following classes are appropriate

Relative freq for =

Class	Frequency	Relative Frequency
45-54	3	3/25 = 0.12
55-64	5	0.2
65-74	8	0.32
75-84	4	0.16
85-94	5	0.2
Total	25	1

b) For making the histogram we need to make the classes continuous. This can be done by adding and subtracting to upper limit and from lower limit respectively

For each class 0.5 has to be added and subtracted. Therefore

Class	Frequency
44.5-54.5	3
54.5-64.5	5
64.5-74.5	8
74.5-84.5	4
84.5-94.5	5
Total	25

 

c) Mean: For mean we need to add an extra column of classmark =

Class	Classmark	Frequency	Classmark * Freq
44.5-54.5	49.5	3	3*49.5= 148.5
54.5-64.5	59.5	5	297.5
64.5-74.5	69.5	8	556
74.5-84.5	79.5	4	318
84.5-94.5	89.5	5	447.5
Total		25	1767.5

Mean =

Mean = 70.7

Q2)

0	0.15	0	0
1	0.2	0.2	0.2
2	0.3	0.6	1.2
3	0.35	1.05	3.15
Total	1	1.85	4.55

For it to be a probability distribution

1. All probabilities should 0 < P < 1: Which is fullfiled
2. Sum of all probabilities should = 1: This condition is also fulfilled.

Therefore this is probability distribution.

Mean =

Mean = 1.85

S.D.=

S.D =

S.D. =

S.D. = 1.0618

Q3)

This is question of binomial distribution. Where X is the event of students working part-time.

Where p = 1 - q

Probability that at least 3 students work part-time = ( 3 or more than 3)

= P ( X = 3) + P( X = 4) + P( X = 5)

=

FinalAns = 0.017909