Question

35. Find the mean of the following data:
Class
200 - 299 300 - 399 400 - 499 500 - 599 600 - 699 700 - 799 800 - 899
Frequency
3
61
118
139
126
151
2​

Answer 1

Answer:

Here, the class- intervals are formed by exclusive method. If we make the series an inclusive one the mid-values remain same. So, there is no need to convert the series into an inclusive form.

Let the assumed mean be A= 749.5 and h= 100.

Calculation of Mean

Life time (in hrs): Frequency

f

i

Mid-Values

x

i

d

i

=x

i

−A=x

i

−749.5 u

i

=

h

x

i

−A

u

i

=

100

x

i

−749.5

f

i

u

i

300-399 14 349.5 -400 -4 -56

400-499 46 449.5 -300 -3 -138

500-599 58 549.5 -200 -2 -116

600-699 76 649.5 -100 -1 -76

700-799 68 749.5 0 0 0

800-899 62 849.5 100 1 62

900-999 48 949.5 200 2 96

1000-1099 22 1049.5 300 3 66

1100-1199 6 1149.5 400 4 24

N=∑f

i

=400 ∑f

i

u

i

=-138

We have,

N= 400, A= 749.5, h= 100 and ∑f

i

u

i

=−138

∴

X

=A+h[

N

1

∑f

i

u

i

]

⇒

X

=749.5+100(

400

−138

)=749.5−34.5=715.

Hence, the average life time of a tube is 715 hours.

Answer 2

Answer:

The mean of the data is $580.33$.

Step-by-step explanation:

Data given,

Class                     Frequency

$200 - 299$                  $3$

$300 - 399$                  $61$

$400 - 499$                 $118$

$500 - 599$                 $139$

$600 - 699$                 $126$

$700 - 799$                 $151$

$800 - 899$                   $2$

To find: The mean of the data =?

• Mean is defined as the sum of the observations divided by the total number of observations.
• Mean = $\frac{\sum x_{i}f_{i} }{\sum f_{i} }$

Here,

• $x_{i}$ = The midpoint of the class
• $f_{i}$ = The frequency

For this, we have to create a table with the midpoint and the product of the frequency and midpoint.

Class             Frequency($f_{i}$)       Midpoint ($x_{i}$)         $x_{i}f_{i}$

$200 - 299$              $3$                        $249.5$                    $748.5$

$300 - 399$              $61$                       $349.5$                   $21319.5$

$400 - 499$            $118$                       $449.5$                    $53041$  

$500 - 599$             $139$                       $549.5$                  $76380.5$

$600 - 699$             $126$                       $649.5$                   $81837$

$700 - 799$              $151$                       $749.5$                 $113174.5$

$800 - 899$               $2$                         $849.5$                   $1699$

Total                   600                                              348200

Now, after putting the values of $x_{i}f_{i}$ and $f_{i}$ in equation (1), we get:

• Mean = $\frac{\sum x_{i}f_{i} }{\sum f_{i} }$
• Mean = $\frac{348200}{600}$
• Mean = $580.33$.

Hence, the mean of the data is $580.33$.