For a data consisting of $15$ observations $x_i$,$i=1, 2, 3, \ldots, 15$,the following results are obtained: $\sum_{i=1}^{15} x_i = 170$ and $\sum_{i=1}^{15} x_i^2 = 2830$. If one of the observations,namely $20$,was found to be wrong and was replaced by its correct value $30$,then the corrected variance is:
- A$80$
- B$78$
- C$76$
- D$75$
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The variance of the first $n$ natural numbers is
Medium
View SolutionIf the mean and variance of five observations are $\frac{24}{5}$ and $\frac{194}{25}$ respectively,and the mean of the first four observations is $\frac{7}{2}$,then the variance of the first four observations is equal to:
DifficultJEE MAIN 2024
View SolutionThe arithmetic mean and standard deviation of a data set of nine numbers are $13$ and $5$ respectively. If $3$ is included as the $10^{th}$ item of the data,then the variance of the data set of ten numbers is:
The mean and standard deviation of $15$ observations are found to be $8$ and $3$ respectively. On rechecking,it was found that,in the observations,$20$ was misread as $5$. Then,the correct variance is equal to......
An analysis of monthly wages paid to workers in two firms $A$ and $B$,belonging to the same industry,gives the following results:
\text{Particulars} \text{Firm } $A$ \text{ and Firm } $B$ \text{No. of wage earners} $A: 586, B: 648$ \text{Mean of monthly wages} $A: Rs. 5253, B: Rs. 5253$ \text{Variance of wages} $A: 100, B: 121$
Which firm,$A$ or $B,$ shows greater variability in individual wages?
| \text{Particulars} | \text{Firm } $A$ \text{ and Firm } $B$ |
| \text{No. of wage earners} | $A: 586, B: 648$ |
| \text{Mean of monthly wages} | $A: Rs. 5253, B: Rs. 5253$ |
| \text{Variance of wages} | $A: 100, B: 121$ |
Which firm,$A$ or $B,$ shows greater variability in individual wages?
Medium
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