Question

Find the missing frequencies in the following frequency distribution if it is known that the mean of the distribution 50.
x: 10 30 50 70 90
y: 17 f1 32 f2 19

Answer 1

The missing frequencies in the given frequency distribution are $f_1=28$ and $f_2=24$

Step-by-step explanation:

Given that the following frequency distribution

x: 10   30    50     70    90

y: 17    $f_1$      32      $f_2$     19

Also given that Mean=50 and N=120

To find the missing frequencies  in the given frequency distribution :

x         y               xy

10         17                 170

30        $f_1$                $30f_1$

50        32               1600

70         $f_2$                $70f_2$

90         19                1710

___________________________

      N=120      $\sum xy=3480+30f_1+70f_2$

____________________________

Since N=120=sum of the frequencies

• $N=17+f_1+32+f_2+19=120$
• $68+f_1+f_2=120$
• $f_1+f_2=120-68$

$f_1+f_2=52\hfill (1)$

Since mean=50

We have that $Mean=\frac{\sum xy}{N}$

• $50=\frac{3480+30f_1+70f_2}{120}$
• $50(120)=3480+30f_1+70f_2$
• $6000=3480+30f_1+70f_2$
• $-30f_1-70f_2=3480-6000$
• $-30f_1-70f_2=-2520$

• Dividing by -10 on both sides we get

$3f_1+7f_2=252\hfill (2)$

• Now solving the equations (1) and (2)
• Multiply equation (1) into 3 we get
• $3f_1+3f_2=156\hfill (3)$

Now subtracting the equations (2) and (3)

$3f_1+7f_2=252$

$3f_1+3f_2=156$

(-)__(-)___(-)________

$4f_2=96$

$f_2=\frac{96}{4}$

Therefore $f_2=24$

Substitute the value of $f_2$ in equation (1) we get

$f_1+24=52$

$f_1=52-24$

Therefore $f_1=28$

Therefore the missing frequencies are $f_1=28$ and

$f_2=24$