Find Coefficient of Variance for the values given : {13,35,56,35,77}
a. 0.4156
b. 0.5164
c. 0.5614
d. 0.6514
Ans  c
Explanation :
Number of terms (N) = 5
Mean:
Xbar = (13+35+56+35+77)/5
= 216/5
= 43.2
Standard Deviation (SD):
Formula to find SD is
σx= √(1/(N  1)*((x1xm)2+(x2xm)2+..+(xnxm)2))
=√(1/(51)((1343.2)2+(3543.2)2+(5643.2)2+(3543.2)2+(7743.2)2))
=√(1/4((30.2)2+(8.2)2+(12.9)2+(8.2)2+(33.8)2))
=√(1/4((912.04)+(67.24)+(163.84)+(67.24)+(1142.44)))
=√(588.2)
=24.2528
Coefficient of variation (CV)
CV = Standard Deviation / Mean
= 24.2528/43.2
= 0.5614
Hence the required Coefficient of Variation is 0.5614
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A bank calculates that its individual savings accounts are normally distributed with a mean of Rupees 2,000 and a standard deviation of Rupeess600. If the bank takes a random sample of 100 accounts, what is the probability that the sample mean will lie between Rupees 1,900 and Rupees 2,050?
a. 0.4792
b. 0.7492
c. 0.7942
d. 0.9742
Ans  b
Explanation :
Standard Error = SD / √(N)
= 600 / √100
= 600 / 10
= 60
Using the equation
z = (x bar minus Mu)/SE
we get 2 z values
for x bar = Rs. 1900,
z = (1900  200) / 60
= (100) / 60
= 1.67
for x bar = Rs. 2050,
z = (2050  200) / 60
= 50 / 60
= 0.83
Probability table gives us probability of 0.4525 corresponding to a z value of –1.67, and it gives probability of 0.2967 for a z value of 0.83. If we add these two together, we get 0.7492 as the total probability that the sample mean will lie between Rs. 1900 and Rs. 2,050.
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Suppose that a population with N is equal to 144 has p is equal to 24. What is the mean of the sampling distribution of the mean for samples of size 25?
a. 24
b. 12
c. 4.8
d. 2
Ans – a
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We have six students say A, B, C, D, E, F participating in a quiz contest. Out of six students only two can reach to the final. What is the probability of reaching to the final of each student?
a. 2/5
b. 1/2
c. 1/3
d. 1/4
Ans  c
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Suppose a population with N = 144 has u(Mean)=24. What is the mean of sampling distribution of the mean for samples of size of Rs 25 ?
a. 24
b. 2
c. 4.8
d. 3.2
Ans  a
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