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Answer Sheet
Mid Term – QT
Answer 1 Sub – Divided Bar chart is constructed in excel sheet and sent separately.
Answer 2 Solution Values
Mid-point
X
f
X-22.5 ord
fd`
fd`2
cf
5-10
10-15
15-20
20-25
7.5
12.5
17.5
22.5
5
10
15
20
-15
-10
-5
0
-15
-20
-15
0
45
40
15
0
5
15
30
50
25-30
30-35
35-30
27.5
32.5
37.5
15
10
5
N=80
5
10
15
15
20
15
∑fd`= 0
15
40
45
200
65
75
80
Karl Pearson’s Coefficient of Skewness = 3 (Mean – Median) / Std. Deviation
Mean = A + ∑fd`/N * C = 22.5 + 0/80 * 5 = 22.5
Median is the size of N/2 or 80/2 or 40th Item.
Thus median lies in class 20-25
By interpolation:
Median = l1 + (N/2 – cf) / f * C = 20 + (40-30)/20 * 5 = 22.5
Std. Deviation of σ = √ [ ∑fd`2/N ] – { ∑ fd`/N}2 * C
= √ 200/80 – {0/80)2 * 5
= √ 2.5 * 5 = 1.6 * 5 = 8
Coefficient of Skewness = 3 (Mean – Median) /Std. Deviation
= 3 (22.5 – 22.5) / 8 = 0
Karl Pearson’s Coefficient of Skewness = 0
Since Coefficient of Skewness is 0, the distribution is symmetrical.
Answer 3 –
Solution A –
Effective as success = p
and non-effective as failure = q
p = 0.4 (since the drug is effective 40% of the time)
q = (1-p) = (1-0.4) = 0.6
x = 2
n = 4
Then: P (x) = ( n ) (p)2 (q)2
x
= ( 4 ) (0.4)2 (0.6)2
2
= 4! / 2! 2! (0.4)2 (0.6)2
= 6 * 0.16 * 0.36
= 0.3456
The mean and the standard deviation for the binomial distribution are given as follows:
µ = np = 4*0.4 = 1.6
and,
σ = √ npq = √ 4*0.4*0.6 = √ 0.96 = 0.979
Solution C
Let A be the event that student passes in statistics and B the event that student passes in
Mathematics test. We are given
P (A) = 2/3 , P (A ∩ B) = 14/45, and P (A U B) = 4/5 and
We want P (B) :
P (A U B) = P (A) + P (B) - P (A ∩ B)
= 4/5
= 2/3 + P (B) – 14/45
= P (B) = 4/5 + 14/45 – 2/3 = 36+14 – 30/ 45
= 20/45 = 4/9
Answer 4 –
Solution –
The population mean µ = ∑X / N = 2 + 4 + 6 + 8 + 10 /5 = 30/ 5 = 6
µ=6
Standard Deviation σ = √ ∑ (X - µ) 2 / N
X
2
4
6
8
10
µ
6
6
6
6
6
(X - µ) 2
16
4
0
4
16
= ∑ (X - µ) = 40
Then σ = √ 40/5 = √ 8 = 2.83
σ = 2.83