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Note on Arithmetic Mean

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The word "statistics " refers two meanings. In the singular sense, it deals with the collection, presentation, analysis and interpretation of numerical data and helps in making a decision. In the plural sense, it refers to the numerical facts and figures are sometimes known as statistical data.

Measures of central tendency

.

A typical value which represents the characteristics of the entire mass of huge data is called the central value of the whole distribution. A measure of the central tendency is also known as a measure of location or an average.
The various measure of central tendency are as follow:

  1. Arithmetic Mean (average)
  2. Median
  3. Mode

Arithmetic Mean

The arithmetic mean is defined as the total sum of observations divided by the total number of observations. Here we study about the arithmetic mean of continuous series.

Calculation of arithmetic means in continuous series

When the number of items is large, we have to divide them into groups. Such groups are known as simple classes. It should be noted that there is no gap between any two successive intervals. So the data is continuous and the series is called continuous series.

Arithmetic Mean can be calculated in three different methods.

  1. \(Direct \: method, \overline{X} = \frac{\sum fm}{n}\), where m is the mid-value of class intervals.
  2. \(Short \: cut \: method,\overline{X} = A + \frac{\sum fd}{n} \), where d= X - A and A is assumed mean.
  3. \(Step \: deviation \: method, \overline{X} = A + \frac{\sum fd'}{n}\) \( \times h \), where d' = \(\frac{d}{h}\) and h is the length of class interval.


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Arithmetic Mean can be calculated in three different methods.

  1. \(Direct \: method, \overline{X} = \frac{\sum fm}{n}\), where m is the mid-value of class intervals.
  2. \(Short \: cut \: method,\overline{X} = A + \frac{\sum fd}{n} \), where d= X - A and A is assumed mean.
  3. \(Step \: deviation \: method, \overline{X} = A + \frac{\sum fd'}{n}\) \( \times h \), where d' = \(\frac{d}{h}\) and h is the length of class interval.
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Very Short Questions

Solution:

Mean(\(\overline X\))= 5o
\(\sum fx = 750 \)
N = ?

\begin{align*} Mean\:(\overline {X}) &= \frac{\sum fx}{N}\\ 50 &= \frac{750}{N}\\ or, N &= \frac{750}{50}\\ \therefore N &= 15 \: \: \: \: _{Ans} \end{align*}

Solution:

\(Mean \: (\overline{X}) =60 \\ \sum fx = 960 \\ Number \: of \: term \: (N) = ? \)

\begin{align*} Mean (\overline {X}) &= \frac{\sum fx}{N}\\ 60 &= \frac{960}{N}\\ or, N &= \frac{960}{60}\\ \therefore N &= 16 \: _{Ans} \end{align*}

Solution:

\(Mean \: (\overline{X}) =12 \\ \sum fx = 70 +10a\\ No. \: of \: term \: (N) = 5+a \)

\begin{align*}Mean \: (\overline{X}) &= \frac{\sum fx}{N}\\ 12 &= \frac{70 + 10a}{5 + a}\\ or, 60 +12a &=70 + 10a\\ or,12a -10a &= 70 - 60\\ or, 2a &= 10 \\ or, a &= \frac{10}{2}\\ \therefore a &= 5 \: \: _{Ans} \end{align*}

Solution:

\(Mean \: (\overline{X}) = 13 \\ No. \: of \: terms \: (N) = 5 + a \)

\begin{align*} Mean (\overline{X}) &= \frac{\sum X}{N} \\ or, 13 &= \frac{4 + 8+12+x+25}{5}\\ or, 65 &=49 + x\\ or, x &= 65 - 49 \\ x &= 16 \:\: _{Ans} \end{align*}

Solution:

First item mean \( (\overline {X_{1}}) = 7 \)
Number of first item \(n_{1} = 4 \)
Second item mean \( (\overline {X_{2}}) = 12\)
No. of Second item \(n_{2}\)= 3
Mean of 7 items \(\overline{X}= ?\)

\begin{align*} ( \overline {X}) &= \frac{n_1 \overline{X}_1 + n_2 \overline{X}_2}{n_1+n_2}\\ &= \frac{4 \times 7 + 3 \times 12}{4 + 3}\\ &= \frac{28 + 36}{7}\\ &= \frac{64}{7}\\ &= 9.14 \: \: \: _{ans} \end{align*}

\(Mean (\overline{X}) = 40 \\ No. \: of \: terms (N)= 7 \\ k = ? \)

\begin{align*} Mean(\overline{X}) &= \frac{\sum X}{N} \\ or, 40 &= \frac{10+20+30+40+50+ 60+30+k}{7}\\ or, 280 &= 240 + k\\ or, k &= 280 - 240\\ \therefore k &= 40 \: \: \: \: \: \: \: \: \: _{Ans} \end{align*}

Solution:

Expenditure (in Rs.)x Frequency(f) fx
24 2 48
25 4 100
30 3 90
35 4 140
40 2 80

\begin{align*} Mean(\overline{X}) &= \frac{\sum fx}{N}\\ &= \frac{458}{15}\\ &= 30.53 \\ \therefore Average \: expenditure &= Rs \: 30.53 \end{align*}

Solution:

Let, the age of remaining students be x.

\(average \: age (\overline{X}) = 9 \\ Number \: of \: terms (N) = 5 \)

\begin{align*} (\overline{X}) &= \frac{\sum X}{N}\\ or, 9 &= \frac{5+7+8+15+x}{5}\\ or, 45 &= 35 + x \\ \therefore x &=45 -35 = 10 \\ \: \\ \therefore & \text{The age of remaining student is 10} \end{align*}

Solution:

Calculation of mean

Class interval cf. f mid-value (m) fm
10-20 4 4 15 60
20-30 16 12 25 300
30-40 56 40 35 1400
40-50 97 41 45 1845
50-60 124 27 55 1485
60-70 137 13 65 845
70-80 146 9 75 675
80-90 150 4 85 340
N = 150 \(\sum fm=6950\)

\begin{align*}Mean \: (\overline{X})&= \frac{\sum fm}{N}\\ &= \frac{6950}{150}\\ &= 46.33 \: _{Ans} \end{align*}

Solution:

Calculation of mean

class interval midvalue f fm
0-10 5 5 25
10-20 15 7 105
20-30 25 8 200
30-40 35 4 140
40-50 45 6 270
N= 30 \( \sum fm = 740 \)

\begin{align*} Mean (\overline{X}) &= \frac{\sum fm}{N}\\ &= \frac{740}{30}\\ &= 24.6 \: _{ans} \end{align*}

Solution:

To find the value of p.

Solution:

To find the value of p.

Age in year No. of teachers (f) m fm
10-20 3 15 45
20-30 8 25 200
30-40 15 35 525
40-50 p 45 45p
50-60 4 55 220
N=p+30 \( \sum fm = 45p + 990\)

We know that,

\begin{align*} Mean \: (\overline {X}) &= \frac{\sum fm}{N}\\ 36 &= \frac{45p + 990}{p+30}\\ or, 1080 + 36p &= 45p +990\\ or, 45p - 36p &= 1080 -990 \\ or, 9p &= 90 \\ \therefore p &= \frac{90}{9} = 10 \: _{Ans} \end{align*}

Solution:

To find the value of p.

Solution:

To find the value of p.

Age in year No. of teachers (f) m fm
10-20 3 15 45
20-30 8 25 200
30-40 15 35 525
40-50 p 45 45p
50-60 4 55 220
N=p+30 \( \sum fm = 45p + 990\)

We know that,

\begin{align*} Mean \: (\overline {X}) &= \frac{\sum fm}{N}\\ 36 &= \frac{45p + 990}{p+30}\\ or, 1080 + 36p &= 45p +990\\ or, 45p - 36p &= 1080 -990 \\ or, 9p &= 90 \\ \therefore p &= \frac{90}{9} = 10 \: _{Ans} \end{align*}

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  • The mean of 50 observations was 250. It was detected on checking that  the value of 165 was wrongly copied as 115 for computation of mean. Find the correct mean.

    251


    345


    152


    543


  • The mean of 40 observations was 160. It was detected on rechecking that the value of 165 was wrongly copied as 125 for computation of mean. Find the correct mean.

    785


    465


    321


    161


  • In a grouped data, the sum of the marks obtained by 50 students in mathematics is 3000.Find the average  mark.

    20 


    80


    60


    60


  • In a grouped data, the sum of the marks obtained by 50 students in mathematics is 3000.Find the average  mark.

    60


    60


    80


    20 


  • In a continuous series the value of a assumed mean is 35 and the sum of frequencies is 50.If mean of the data is 41 then find the sum of the product of deviation (d) and frequencies(f).

    700


    750


    200


    300


  • In a continuous series the value of a assumed mean is  25 and the sum of frequencies is 150.If mean of the data is 26.07  then find the sum of the product of deviation (d) and frequencies(f).

    120


    80


    175


    160


  • If the mean of a grouped data having ∑fm =306 is 18, find the value of N.

    10


    17 


    7


    20


  • In a continuous frequency distribution data, the number of terms (N)  = 40 and mean ((overline {X}) )= 75. If a class interval 40-50 having frequency 10 is included in the series, find the new mean.

    92


    69


    22


    54


  • In a continuous frequency distribution data, the number of terms (N)  = 40 and mean (overline {X}) )= 75. If a class interval 70-80 having frequency 12 is excluded from  the series, what will be the new mean?

    75
    85
    40
    55
  • In a continuous frequency distribution data, the number of terms (N)  = 40 and mean ((overline {X}) )= 36. If a class interval 40-50 having frequency 10 is excluded from  the series, what will be the new mean?

    21


    65


    45


    33


  • In a continuous series the mean of 50 observation was 40. It was detected on checking that the class interval 30-40 having frequency 10 was wrongly copied as class interval 30-50 having frequency 10 for computation of mean.Find the correct mean.

    69


    49


    59


    39


  • In a continuous series the mean of 50 observation was 40. It was detected on checking that the class interval 30-40 having frequency 10 was wrongly copied as class interval 30-50 having frequency 10 for computation of mean.Find the correct mean.

    39


    69


    59


    49


  • In a continuous series the mean of 40 observation was 60. It was detected on checking that the class interval 20-40 having frequency 10 was wrongly copied as class interval 20-40 having frequency 10 for computation of mean.Find the correct mean.

    61.25


    35.30


    31


    40.25


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Niva

In the continuos series mean=15 a and £fx = 420 a find the number of terms(N)


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Ask any queries on this note.find the value of meanmark 0-10 0-20 0-30 0-40 0-50f 5 10 14 17 20


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