In a factory 50 workers are employed the monthly salaries of 50 workers of a company are show in the table salaries in 1000s 12-14,14-16,16-18,18-20,20-22,22-24 frequency 6,8,12,7,9,8 1.what is the estimated mean of the salary?2.what is the median of the salary?3.a worker is selected at rabdom what is the probabilitythat his salary less than 16000?4.more 16000?5.how many workers earn more money than the mean salary?
Answers
Solution :-
Salary(in 1000) --------Fi ------- Xi --------- FiXi ------ CF
12 - 14 -------------------- 6 ------- 13 ---------- 78 ------- 6
14 - 16 -------------------- 8 ------- 15 ---------- 120 -------14
16 - 18 --------------------12 ------- 17 ---------- 204 -------26
18 - 20 --------------------7 ------- 19 ---------- 133 ------- 33
20 - 22 --------------------9 -------21 ---------- 189 ------- 42
22 - 24 --------------------8 ------- 23 ----------184 ------- 50
⅀Fi = 50 ⅀FiXi = 908
so,
→ Mean salary = ⅀FiXi / ⅀Fi = 908000 / 50 = Rs.18,160 .
now, Here , n = 50 .
So,
→ (n/2) = 25.
Then,
- Cumulative frequency greater than 25 is 26, corresponds to the class 16 - 18.
Therefore,
→ Class 16 - 18 is the median class.
Now,
- Median = l + [{(n/2) - cf} / f] * h
from data we have :-
- l = lower limit of median class = 16000
- n = total frequency = 50
- cf = Cumulative frequency of class before median class = 14
- f = frequency of median class = 12 .
- h = size of class = 2000 .
Putting all value we get :-
→ Median = 16000 + [(25 - 14)/12] * 2000
→ Median = 16000 + (11/12) * 2000
→ Median = 16000 + 1833.34
→ Median = 17833.34 .
now,
→ Probability of salary less than 16000 = 14/50 = (7/25) .
→ Probability of salary more than 16000 = 36/50 = (18/25) .
→ Number of workers having salary more than mean salary = 7 + 9 + 8 = 24 .
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