Q. 103.7( 3 Votes )

# The following distribution shows the number of out door patients in 64 hospitals as follows. If the mean is 18, find the missing frequencies f_{1} and f_{2} :

Answer :

This is a grouped frequency distribution.

To find f_{1} and f_{2}, we’ll need to find mean of the following distribution by direct method and equate it to the given mean of the following distribution, 18.

This given data is exclusive, grouped frequency distribution.

So, let’s construct a table finding midpoints and stating frequencies.

So now, we have

∑x_{i}f_{i} = 608 + 16f_{1} + 20f_{2}

And ∑f_{i} = 35 + f_{1} + f_{2}

Mean is given by

⇒ [given, mean = 18 and using the values from the table]

⇒ 18 × (35 + f_{1} + f_{2}) = 608 + 16f_{1} + 20f_{2}

⇒ 630 + 18f_{1} + 18f_{2} = 608 + 16f_{1} + 20f_{2}

⇒ 20f_{2} – 18f_{2} + 16f_{1} – 18f_{1} = 630 – 608

⇒ 2f_{2} – 2f_{1} = 22

⇒ 2(f_{2} – f_{1}) = 22

⇒ f_{2} – f_{1} = 11 …(i)

Since, there are 64 hospitals.

⇒ Number of hospitals = 64

⇒ Frequency = 64

⇒ ∑f_{i} = 64

⇒ 35 + f_{1} + f_{2} = 64

⇒ f_{1} + f_{2} = 64 – 35

⇒ f_{1} + f_{2} = 29 …(ii)

Solving equations (i) and (ii), we get

f_{2} – f_{1} = 11

f_{2} + f_{1} = 29

2f_{2} + 0 = 40

⇒ 2f_{2} = 40

⇒

⇒ f_{2} = 20

Putting f_{2} = 20 in equation (ii), we get

f_{1} + 20 = 29

⇒ f_{1} = 29 – 20

⇒ f_{1} = 9

**Thus, the missing frequencies are f _{1} = 9 and f_{2} = 20.**

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