Following are some of the few Statistics Problems from the wide ranging problems in statistics
The distribution below shows the number of wickets taken by bowlers in one-day cricket matches. Find the mean number of wickets by choosing a suitable method. What does it signify?
Number of wickets 20-60 60-100 100-150 150-250 250-350 350-450
Number of bowlers 7 5 16 12 2 3
Solution: the class size varies and the x(i)’s are large. We shall apply the step deviation method with a=200 and h=20. Let us tabulate the values
No. of wickets Number of x(i) d(i)=x(i)-200 u(i)=d(i)/20 u(i)f(i)
Taken bowlers
20-60 7 40 -160 -8 -56
60-100 5 80 -120 -6 -30
100-150 16 125 -75 -3.75 -60
150-250 12 200 0 0 0
250-350 2 300 100 5 10
350-450 3 400 200 10 30
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Total 45 -106
So, u(bar) = -106/45 and hence, x(bar) = 200 + 20[-106/45]= 200 – 47.11 = 152.89
This tells us that, on an average, the number of wickets taken by these 45 bowlers in one-day cricket is 152.89
Let us solve Statistics problems involving confidence interval; A sample of 16 students is taken. The average age in the sample was 22 years with the standard deviation of 6 years. Construct a 95% confidence interval for the average age of the population
Solution: The formula to construct a 95% confidence interval for the mean population can be given as,
C.I. = [{Xn(bar)+z(alpha/2) sigma/sqrt(n)} , { Xn(bar)-z(alpha/2) sigma/sqrt(n)}]
Where Xn(bar)=22 is the sample mean, sigma=6 is the standard deviation given, n=16 is the sample size, z(alpha/2) is the cutoff point to the standard normal deviation. Since we want a 95% confidence interval , we consider alpha =0.05 and with the help of the normal distribution table we can arrive to z(alpha/2) = 1.96. Plugging in all the data we have in the formula, we get
C.I = {[22+1.96(6/sqrt(16)], [22-1.96(6/sqrt(16)]= (24.94, 19.06) is the required confidence interval
Statistic problem to find the variance and variation for the data given, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24
In this problem, the mean is calculated by step-deviation method taking 14 as the assumed mean. The number of observations n=10. Here the x(i) values are 6,8,10,12,14,16,18,20,22,24; d(i) values we get are, -4, -3,-2,-1, 0, 1,2,3,4,5 and summation of d(i) is 5. The deviation of mean [x(i)- x(bar)] values are -9, - 7,-5,-3, -1, 1, 3,5, 7, 9 and finally the [x(i)- x(bar)] are 81, 49, 25, 9, 1, 1, 9, 25, 49,81, summation of which is 330. Mean is calculated using the formula,
Mean x(bar) = assumed mean + [summation(i=1ton) d(i) *h]/n
= 14 + (5/10)x 2 = 15
Variance (sigma^2) = 1/n summation (i=1to10)[x(i)-x(bar)]^2 = (1/10) x 330 = 33
Standard deviation = sqrt(sigma) = sqrt(33) = 5.74
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