$A$ factory is operating in two shifts,day and night,with $70$ and $30$ workers respectively. If the per day mean wage of the day shift workers is $Rs. 54$ and the per day mean wage of all the workers is $Rs. 60$,then the per day mean wage of the night shift workers (in $Rs.$) is:
- A$69$
- B$66$
- C$74$
- D$75$
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Similar Questions
For a group of $100$ observations,the arithmetic mean and standard deviation are $8$ and $\sqrt{10.5}$ respectively. The mean and standard deviation of $50$ items selected from these $100$ observations are $10$ and $2$ respectively. Then the standard deviation of the remaining $50$ observations is
If the coefficients of variation of two distributions are $40$ and $20$ and their variances are $144$ and $64$ respectively,then the mean of their arithmetic means is
The mean of $n$ observations $x_{1}, x_{2}, \ldots, x_{n}$ is $\bar{x}$. If an observation $x_{q}$ is replaced by $x_{q}^{\prime}$, then the new mean is:
An analysis of monthly wages paid to workers in two firms $A$ and $B$,belonging to the same industry,gives the following results:
\text{Parameter}
\text{Firm } $A$ \text{ and Firm } $B$
\text{No. of wage earners}
$A: 586, B: 648$
\text{Mean of monthly wages}
$Rs. 5253$
\text{Variance of wages}
$A: 100, B: 121$
Which firm,$A$ or $B$,pays a larger total amount as monthly wages?
| \text{Parameter} | \text{Firm } $A$ \text{ and Firm } $B$ |
| \text{No. of wage earners} | $A: 586, B: 648$ |
| \text{Mean of monthly wages} | $Rs. 5253$ |
| \text{Variance of wages} | $A: 100, B: 121$ |
Which firm,$A$ or $B$,pays a larger total amount as monthly wages?
Medium
View SolutionThe mean of $6$ distinct observations is $6.5$ and their variance is $10.25$. If $4$ out of $6$ observations are $2, 4, 5$ and $7$,then the remaining two observations are:
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