Question
Download Solution PDFIf mean of the following distribution is 211, then what is the value of (4p - 3q) ?
Class | 100 - 150 | 150 - 200 | 200 - 250 | 250 - 300 | 300 - 350 | Total |
Frequency | 4 | p | 12 | q | 2 | 25 |
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFFormula used:
Mean (x̅) = (Summation of fi xi) / (Summation of fi)
Where xi is the class mark and fi is the frequency.
Class Mark (xi) = (Lower Limit + Upper Limit) / 2
Calculation:
First, calculate the class mark (xi) for each class and then fixi:
Class | Frequency (fi) | Class Mark (xi) | fixi |
100 - 150 | 4 | (100+150)/2 = 125 | 4 × 125 = 500 |
150 - 200 | p | (150+200)/2 = 175 | p × 175 = 175p |
200 - 250 | 12 | (200+250)/2 = 225 | 12 × 225 = 2700 |
250 - 300 | q | (250+300)/2 = 275 | q × 275 = 275q |
300 - 350 | 2 | (300+350)/2 = 325 | 2 × 325 = 650 |
Total | 25 | 500 + 175p + 2700 + 275q + 650 = 3850 + 175p + 275q |
From the total frequency, we have:
4 + p + 12 + q + 2 = 25
⇒ 18 + p + q = 25
⇒ p + q = 25 - 18
⇒ p + q = 7 ......(Equation 1)
Given Mean = 211
x̅ = (Summation of fixi) / (Summation of fi)
⇒ 211 = (3850 + 175p + 275q) / 25
⇒ 211 × 25 = 3850 + 175p + 275q
⇒ 5275 = 3850 + 175p + 275q
⇒ 5275 - 3850 = 175p + 275q
⇒ 1425 = 175p + 275q
⇒ 1425/25 = 175p/25 + 275q/25
⇒ 57 = 7p + 11q ......(Equation 2)
From Equation 1, p = 7 - q.
Substitute p in Equation 2:
57 = 7(7 - q) + 11q
⇒ 57 = 49 - 7q + 11q
⇒ 57 - 49 = 4q
⇒ 8 = 4q
⇒ q = 8/4
⇒ q = 2
Substitute q = 2 in Equation 1:
p + 2 = 7
⇒ p = 7 - 2
⇒ p = 5
Now, find the value of (4p - 3q):
4p - 3q = 4 × 5 - 3 × 2
⇒ 4p - 3q = 20 - 6
⇒ 4p - 3q = 14
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