Question

3.) The frequency distribution of marks scored by
400 students is as follows. If the mean of the
frequencydistribution is 41.2 marks, find the missing frequencies x and y.
Marks
0-10 10-20 20-30 30 - 40 40-50 50-60 60 – 70 | 70-80
No. of students 26 26 x 110 8 4 y | 36 | 32​

Answer 1

x = 38 and y = 48

Given:

Variable:       0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80

Mid value:       5       15       25      35       45       55       65      75

Frequency:    26,     26,       x,      110,      84,       y,        36,      32

Mean value is 41.2

Total number of items is 400

To find:

Missing frequencies.

Explanation:

Let missing frequency be x.  and y

See the table below attached.

The given mean = 41.2

Mean  = Mid value × frequency ÷ Number of frequency

41.2  = $\frac{5\times 26 +15\times 26 +25\times x +35\times 110+ 45 \times 84+ 55 \times y +65\times 36+ 75\times 32}{400}$

3590 = 25x + 55y    

Total number of items is 400 = 26 + 26 + x + 110+ 84 + y + 32 +36

                                       400 =314 +x+y

                                       86  = x+y

From solving the above equation x= 38 and y= 48

To learn more....

1. The following is the distribution of height of students of certain class in a certain city: Find the average height of maximum number of students.

Height(in cms): 160-162 163-165 166-168 169-171 172-174

No.of students: 15 118 142 127 18

brainly.in/question/15924910

2. The following is the distribution of height of students of certain class in a certain city: Find the average height of maximum number of students.

Height(in cms): 160-162 163-165 166-168 169-171 172-174

No.of students: 15 118 142 127 18

brainly.in/question/15924910

Answer 2

Answer:

X=38

Y=48

Step-by-step explanation:

The Whole Method is given in the Picture .

Thank you

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