In an experiment with $15$ observations on $x$,we have $\sum x^2 = 2830$ and $\sum x = 170$. One observation that was $20$ was found to be wrong and was replaced by the correct value $30$. The corrected variance is:
- A$177.33$
- B$188.66$
- C$80.33$
- D$78$
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The 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:
Two plants $A$ and $B$ of a factory show the following results regarding the number of workers and the wages paid to them:
\text{Parameter} \text{Plant } $A$ \text{ and } $B$ \text{ Data} \text{No. of workers} $A: 500, B: 6000$ \text{Average monthly wages} $A: Rs. 2500, B: Rs. 2500$ \text{Variance of distribution of wages} $A: 81, B: 100$
In which plant,$A$ or $B$,is there greater variability in individual wages?
| \text{Parameter} | \text{Plant } $A$ \text{ and } $B$ \text{ Data} |
| \text{No. of workers} | $A: 500, B: 6000$ |
| \text{Average monthly wages} | $A: Rs. 2500, B: Rs. 2500$ |
| \text{Variance of distribution of wages} | $A: 81, B: 100$ |
In which plant,$A$ or $B$,is there greater variability in individual wages?
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View SolutionThe variance for the first six prime numbers greater than $5$ is
If $X$ is a random variable such that $\sigma(X) = 2.6$, then $\sigma(1 - 4X)$ is equal to:
The raw data $x_1, x_2, \ldots, x_{n}$ is an $A$.$P$. with common difference $d$ and first term $0$. If $\bar{x}$ and $\sigma^2$ are the mean and variance of $x_{i}, i=1, 2, \ldots, n$,then $\sigma^2$ is:
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