Class Interval |
Mid-point of Interval |
Count |
Relative Frequency |
Cumulative Relative Frequency |
248-547 |
365.72 |
18 |
37.50% |
37.50% |
548-847 |
662.67 |
6 |
12.50% |
50.00% |
848-1147 |
985.22 |
9 |
18.75% |
68.75% |
1148-1447 |
1271.43 |
7 |
14.58% |
83.33% |
1448-1747 |
1601.00 |
2 |
4.17% |
87.50% |
1748-2047 |
1848.00 |
2 |
4.17% |
91.67% |
2048-2347 |
2268.00 |
1 |
2.08% |
93.75% |
2648-2947 |
2830.00 |
1 |
2.08% |
95.83% |
2948-3247 |
2958.00 |
1 |
2.08% |
97.92% |
7448-7747 |
7729.00 |
1 |
2.08% |
100.00% |
Table 1: Frequency Table of Number of Passengers at Peak time
The average of the number of passengers who are present in a train station in Melbourne during the peak times between 7am and 9:29 am is 1062.6875. The value of the median for the raw data was obtained after having sorted the data in ascending order. Then, as the total number of observations is 48, the median corresponds to 24thobservation in the sorted data and is equal to 733. The mode is the observation with highest frequency. Considering the data in raw ungrouped form, the observation 401 had highest frequency of 2. Thus the mode is 401.
Here, n = number of sampled observations, = ith observation in the sample, 1 ≤ i ≤n,
= mean weekly attendance.
The calculations for the sample standard deviation are shown in the following table:
X |
X- |
(X-)2 |
472 |
-12.5714 |
158.0408 |
413 |
-71.5714 |
5122.469 |
503 |
18.42857 |
339.6122 |
612 |
127.4286 |
16238.04 |
399 |
-85.5714 |
7322.469 |
538 |
53.42857 |
2854.612 |
455 |
-29.5714 |
874.4694 |
Table 2: Calculation for Standard Deviation of Weekly attendance
The standard deviation, using the formula specified above, was then computed to be equal to 68.559.
The inter quartile range is the difference of the third quartile from the first quartile. The value of the first quartile was computed 6014. The value of the third quartile was computed as 7223. Thus the inter quartile range was computed as 1209.
The inter quartile range is unaffected by outliers that may be present in the data. Hence unlike standard deviation which depends on the mean and hence is influenced by outliers, the inter quartile range can be particularly useful in measuring variation of a dataset which has extreme values.
The correlation coefficient computed using observations on attendance and the sales of chocolate bar was computed as 0.967. Therefore a strong positive relation exists between the two variables.
Let X denote the weekly attendance at Holmes and let Y be the number of chocolate bars that are sold. Taking Y as dependent and X as independent variable, the regression model of the following form is to be determined:
Y= a+ b. X +
The estimate of the parameters after solving the normality equations for solving the linear regression model are:
a =
b=, where, = mean of Y, = mean of X
Computing the value of a and b, table 3 shows the computations:
X |
Y |
X- |
Y- |
2 |
2 |
|
472 |
6916 |
-12.5714 |
113.4286 |
158.0408 |
12866.04082 |
-1425.96 |
413 |
5884 |
-71.5714 |
-918.571 |
5122.469 |
843773.4694 |
65743.47 |
503 |
7223 |
18.42857 |
420.4286 |
339.6122 |
176760.1837 |
7747.898 |
612 |
8158 |
127.4286 |
1355.429 |
16238.04 |
1837186.612 |
172720.3 |
399 |
6014 |
-85.5714 |
-788.571 |
7322.469 |
621844.898 |
67479.18 |
538 |
7209 |
53.42857 |
406.4286 |
2854.612 |
165184.1837 |
21714.9 |
455 |
6214 |
-29.5714 |
-588.571 |
874.4694 |
346416.3265 |
17404.9 |
Table 3: Computations for Regression
The mean of weekly attendance was found to be 484.571, the mean number of chocolate bars sold in a week was found to be 6802.57. Then
b= = 10.977 and a = 6802.57 – 10.977× 484.571 = 1628.6889
The regression model was therefore determined to be:
Y= 1628.6889 + 10.9772X
The sales are expected to increase by 10.9972 units with unit increase in weekly attendance and with zero attendance the expected sales is 1628.6 or 1629 units approximately.
The coefficient of determination or 1-R2 is the proportion of variation explained by the model. It is given by the formula: 1- . The value of the denominator was obtained as 4004031.714. The following table gives the computations done for the um of squared error:
X |
Y |
Predicted Y |
Error( =Y-Predicted Y) |
square of Error |
472 |
6916 |
6668.343 |
-247.657 |
61333.82 |
413 |
5884 |
6038.387 |
154.3866 |
23835.21 |
503 |
7223 |
6999.338 |
-223.662 |
50024.87 |
612 |
8158 |
8163.156 |
5.156081 |
26.58517 |
399 |
6014 |
5888.905 |
-125.095 |
15648.69 |
538 |
7209 |
7373.041 |
164.0408 |
26909.38 |
455 |
6214 |
6486.83 |
272.8304 |
74436.41 |
Table 4: Computations for Coefficient of Determination
The sum of squared error was found to be 252214.9601 Then the coefficient of determination was found to be 1- = 0.937.
Therefore it is seen that the model here explains 93.7% of the total variation in the dependent variable Y, that is the number if chocolate bars sold in a week. This is quite a good fit.
Let A be the event that a student is enrolled in Holmes, let B denote the event that a student is given grass root training, let C be the event that student is an external student and let D be the event that student is given scientific training. The following table gives the observed frequencies of the four events, A, B, C, D.
Scientific training |
Grassroots training |
TOTAL |
|
Recruited from Holmes students |
35 |
92 |
127 |
External recruitment |
54 |
12 |
66 |
TOTAL |
89 |
104 |
193 |
Table 5: Table of Observed frequencies
P (A B) = P (A)+P(B)–P(
Here P(A) and P(B) are the marginal frequencies of “Recruited from Holmes students” and “Grassroots training” divided by total frequency.
P (A) = = 0.658
P (B) = = 0.538
P ( = = 0.476
Then P (A B) =0.658+0.538-0.476 =0.720
P (D|A) =
Here P (D, A) = P( A D) = = 0.181 and P( A) = 0.658 (from part a)
Then P (D|A) = = 0.2775
Here = Expected frequency of ith case, = Observed frequency ith case, 1 ≤ i ≤ 4 (2 cases for each of 2 variable)
The expected cell frequencies computed using the ratio of the product of the row and column totals with that of the total frequency is:
Scientific training |
Grassroots training |
|
Recruited from Holmes students |
58.56476684 |
68.4352332 |
External recruitment |
30.43523316 |
35.5647668 |
Table 6: Expected Cell Frequencies for Chi-squared Test
Then the following table breaks down the computation for the Chi-squared statistic:
Observed(Oi) |
Expected(Ei) |
Oi– Ei |
(Oi– Ei)2 / Ei |
35 |
58.56476684 |
-23.564767 |
9.48178 |
54 |
30.43523316 |
23.5647668 |
18.24524 |
92 |
68.43523316 |
23.5647668 |
8.114216 |
12 |
35.56476684 |
-23.564767 |
15.61372 |
Table 7: Computations for Chi-square Statistics
The chi square statistic was then obtained as per the specified formula as 51.454. The degrees of freedom of the distribution is then (2-1) × (2-1) =1. The critical value of the chi-square statistic is 3.841. Thus the test concludes that the variables are not independent.
Let A denote a person belonging to segment A, similarly let B, C denote a person belonging to segments B,C and D respectively. Again let X denote the phenomenon that a person prefers the product X over Y and Z.
It is given that 20% of the customer belonging to segment A prefers product X over Y and Z, 35% of those in segment B prefers X, 60% of those in segment C prefers X and 90% of those in D prefers X over Y and Z. The conditional probabilities that a person from the segment A, B and C prefer X are:
P(X|A) = 0.2, P(X|B)=0.35, P(X|C)= 0.6, P(X|D) = 0.9
The probabilities of a person belonging to the segments are:
P(A) = 0.55, P(B)=0.33, P(C)=0.1,P(D)=0.05
Then P(A,X) = P(A)P(X|A) = 0.2×0.55 = 0.11
P(B,X) = P(B)P(X|B) = 0.33× 0.35 = 0.105
P(C,X) = P(C )P(X|C) = 0.1× 0.6 = 0.06
P(D,X) = P(D)P(X|D) =0.05× 0.9 =0.045
(b) Then the probability that a person favors X is,
P(X) = P(A,X)+P(B,X)+P(C,X)+P(D,X) = 0.32
P(A|X) = == 0.3437.
The probability that the customer will buy is given to be 1/10. It is also given that 8 customers entered the store in a 1 minute time interval. The number of customers who make a purchase among them then follows a binomial distribution with p being 1/10 and n being 8. Then the probability that 2 out of the 8 make a purchase is given by the probability mass function of Binomial(8, 1/10) distribution, given as:
Then the probability that two of the customers make a purchase is given as,
P(X=2) = = = 0.148.
P(X=x) =, x ≥ 0
Then putting x=9, the probability was found to be = = 0.12
The average price of a Surfers Paradise apartment was determined to be $1.1 million and the standard deviation was found to be $385 000. It is then assumed that the price of a Surfers Paradise apartment follows a normal distribution and that the distribution parameters are specified by mean 1.1 and standard deviation 385000/1000000, that is, 0.358.
Then using the CDF of standard normal, the probability that price is at least $2 million is given by, , being the CDF of standard normal distribution. Using the standard normal table, = = 0.99403. Then the probability that price is greater than $2 million is given by, 1- = 1-0.99403 =0.00597.
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