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.
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:
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.
Arithmetic Mean can be calculated in three different methods.
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*}
Find the mean from the following data.
Class-interval | 10-20 | 10-30 | 10-40 | 10-50 | 10-60 | 10-70 | 10-80 | 10-90 |
Frequency | 4 | 16 | 56 | 97 | 124 | 137 | 146 | 150 |
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*}
Calculate the mean from the following data.
X | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
f | 5 | 7 | 8 | 4 | 6 |
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*}
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.
152
251
543
345
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.
465
161
785
321
In a grouped data, the sum of the marks obtained by 50 students in mathematics is 3000.Find the average mark.
80
20
60
60
In a grouped data, the sum of the marks obtained by 50 students in mathematics is 3000.Find the average mark.
80
60
20
60
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).
750
300
700
200
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).
80
120
175
160
If the mean of a grouped data having ∑fm =306 is 18, find the value of N.
20
17
10
7
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.
69
22
92
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?
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?
45
65
33
21
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.
59
49
39
69
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
59
69
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
40.25
31
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Niva
In the continuos series mean=15 a and £fx = 420 a find the number of terms(N)
Feb 02, 2017
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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
Jan 29, 2017
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