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Standard IX

Mathematics

Mean

If x̅̅1̅ , ...

Question

If $_{1},_{2},_{3}, . . .,_{n}$ are the means of n groups with $n_{1}, n_{2}, . . ., n_{n}$ number of observations respectively, then the mean x of all the groups taken together is given by :

\sum_{i = 1}^{n} n_{i}_{i}

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\frac{\sum_{i = 1}^{n} n_{i}_{i}}{n^{2}}

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\frac{\sum_{i = 1}^{n} n_{i}_{i}}{\sum_{i = 1}^{n} n_{i}}

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\frac{\sum_{i = 1}^{n} n_{i}_{i}}{2 n}

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Solution

The correct option is C $\frac{\sum_{i = 1}^{n} n_{i}_{i}}{\sum_{i = 1}^{n} n_{i}}$
Mean of all the groups taken together = $\frac{T o t a l S u m}{T o t a l C o u n t}$
Mean = $\frac{\sum_{i = 1}^{n} n_{i}_{i}}{\sum_{i = 1}^{n} n_{i}}$

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Similar questions

Q. Question 17
If

_{1},_{2},_{3}, \dots \dots_{n}

, are the means of n groups with

n_{1}, n_{2}, n_{n}

number of observations, respectively. Then the mean

¯ x

of all the groups:

A)

\sum_{i = 1}^{n} n_{i}_{i}

\frac{\sum_{i = 1}^{n} n_{i}_{i}}{n^{2}}

\frac{\sum_{i = 1}^{n} n_{i}_{i}}{\sum_{i = 1}^{n} n_{i}}

\frac{\sum^{n} n_{i}_{i}}{2 n}

If $_{1},_{2} \dots,_{n}$ are the means of n groups with $n_{1}, n_{2}, \dots, n_{n}$ number of observations respectively then the mean $¯ x$ of all the groups taken together is
(a) $\sum_{n}^{i = 1} n,_{i}$

(b) $\frac{\sum_{i = 1}^{n} n_{i}_{i}}{n^{2}}$

(d) $\frac{\sum_{i = 1}^{n} n_{i} {¯ x}_{i}}{2 n}$

Q. If

x_{1}, x_{2}, . . ., x_{n}

are the means of n groups with

n_{1}, n_{2}, . . ., n_{n}

number of observations respectively, then the mean

x

of all the groups taken together is
(a)

\sum_{i = 1}^{n} n_{i} x_{i}

(b)

\sum_{\frac{i = 1}{n^{2}}}^{n} n_{i} x_{i}

(c)

\frac{\sum_{i = 1}^{n} n_{i} x_{i}}{\sum_{i = 1}^{n} n_{i}}

(d)

\frac{\sum_{i = 1}^{n} n_{i} x_{i}}{2 n}

Q. If

{\bar{X}}_{1}, {\bar{X}}_{2}, . . ., {\bar{X}}_{k}

are the means of n group with n1,n2,...,nk number of observations respectively, then the mean

\bar{X}

of all the groups taken together is govern by

(a) \sum_{i = 1}^{k} n_{i} {\bar{X}}_{i} (b) \frac{1}{n^{2}} \sum_{i = 1}^{k} n_{i} {\bar{X}}_{i} (c) \frac{\sum_{i = 1}^{k} n_{i} {\bar{X}}_{i}}{\sum_{i = 1}^{k} n_{i}} (d) \frac{\sum_{i = 1}^{k} n_{i} {\bar{X}}_{i}}{2_{n}}

Q. Assertion :If mean of

7

observation is

10

& mean of set of

3

observation is

5

, then mean of combined set is

8.5

. Reason: If

_{1} &_{2}

be the mean of

n_{1} & n_{2}

observations respectively, then combined mean is

\frac{n_{1}_{1} + n_{2}_{2}}{n_{1} + n_{2}}

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If ¯x1,¯x2,¯x3,...,¯xn are the means of n groups with n1,n2,...,nn number of observations respectively, then the mean x of all the groups taken together is given by :

The correct option is C ∑ni=1ni¯xi∑ni=1niMean of all the groups taken together = TotalSumTotalCountMean = ∑ni=1ni¯xi∑ni=1ni

If $_{1},_{2},_{3}, . . .,_{n}$ are the means of n groups with $n_{1}, n_{2}, . . ., n_{n}$ number of observations respectively, then the mean x of all the groups taken together is given by :

The correct option is C $\frac{\sum_{i = 1}^{n} n_{i}_{i}}{\sum_{i = 1}^{n} n_{i}}$
Mean of all the groups taken together = $\frac{T o t a l S u m}{T o t a l C o u n t}$
Mean = $\frac{\sum_{i = 1}^{n} n_{i}_{i}}{\sum_{i = 1}^{n} n_{i}}$