Statistics Questions

The mean and variance of 5 observations of an experiment are 4 and 5.2 respectively. If from these observations three are 1, 2 and 6, then remaining will be

Solution : As given \(\bar{x}\) = 4, n = 5 and \({\sigma}^2\) = 5.2. If the remaining observations are \(x_1\), \(x_2\) then \({\sigma}^2\) = \(\sum{(x_i – \bar{x})}^2\over n\) = 5.2 \(\implies\) \({(x_1-4)}^2 + {(x_2-4)}^2 + {(1-4)}^2 + {2-4)}^2 + {(6-4)}^2\over 5\) = 5.2 \(\implies\) \({(x_1-4)}^2 + {(x_2-4)}^2\) = 9  …..(1) Also \(\bar{x}\) = 4 \(\implies\) …

The mean and variance of 5 observations of an experiment are 4 and 5.2 respectively. If from these observations three are 1, 2 and 6, then remaining will be Read More »

A student obtained 75%, 80%, 85% marks in three subjects. If the marks of another subject are added then his average marks can not be less than

Solution : Total marks obtained from three subjects out of 300 = 75 + 80 + 85 = 240 if the marks of another subject is added then the total marks obtained out of 400 is greater than 240 if marks obtained in fourth subject is 0 then minimum average marks = \(240\over 400\)\(\times\)100 = …

A student obtained 75%, 80%, 85% marks in three subjects. If the marks of another subject are added then his average marks can not be less than Read More »

If mean of the series \(x_1\), \(x^2\), ….. , \(x_n\) is \(\bar{x}\), then the mean of the series \(x_i\) + 2i, i = 1, 2, ……, n will be

Solution : As given \(\bar{x}\) = \(x_1 + x_2 + …. + x_n\over n\) If the mean of the series \(x_i\) + 2i, i = 1, 2, ….., n be \(\bar{X}\), then \(\bar{X}\) = \((x_1+2) + (x_2+2.2) + (x_3+2.3) + …. + (x_n + 2.n)\over n\) = \(x_1 + x_2 + …. + x_n\over n\) …

If mean of the series \(x_1\), \(x^2\), ….. , \(x_n\) is \(\bar{x}\), then the mean of the series \(x_i\) + 2i, i = 1, 2, ……, n will be Read More »