Question 1191524: 3. Marks of 75 students are summarized in the following frequency distribution:
Marks 40 - 44 45 - 49 50 - 54 55 - 59 60 - 64 65 - 69 70 - 74
No. of students 7 10 22 F4 F5 6 3
If 20% of the students have marks between 55 and 59
i. Find the missing frequencies f4 and f5.
ii. Find the mean, median and mode
Answer by math_tutor2020(3817)   (Show Source):
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Part (i)
Original Table:
| Marks | 40 - 44 | 45 - 49 | 50 - 54 | 55 - 59 | 60 - 64 | 65 - 69 | 70 - 74 | | No. of students | 7 | 10 | 22 | F4 | F5 | 6 | 3 |
20% of all 75 students have marks between 55 and 59.
20% of 75 = 0.20*75 = 15
The number 15 replaces F4
Let x be the value of F5
The items in the frequency column must add to 75
7 + 10 + 22 + 15 + x + 6 + 3 = 75
x+63 = 75
x = 75-63
x = 12
The number 12 replaces F5
Updated Table:
| Marks | 40 - 44 | 45 - 49 | 50 - 54 | 55 - 59 | 60 - 64 | 65 - 69 | 70 - 74 | | No. of students | 7 | 10 | 22 | 15 | 12 | 6 | 3 |
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Part (ii)
The class interval 40 to 44 has the midpoint (A+B)/2 = (40+44)/2 = 84/2 = 42
Through similar calculations, the midpoint of the interval 45 to 49 is 47.
This process is carried out until you address each interval.
The midpoints in order from left to right are: 42, 47, 52, 57, 62, 67, 72
The midpoints are increasing by 5 each time, which is exactly the gap from each left endpoint of adjacent classes (same goes for the right endpoints as well).
I'll add the row of midpoints which I'll refer to as M
| Marks | 40 - 44 | 45 - 49 | 50 - 54 | 55 - 59 | 60 - 64 | 65 - 69 | 70 - 74 | | No. of students | 7 | 10 | 22 | 15 | 12 | 6 | 3 | | Midpoint (M) | 42 | 47 | 52 | 57 | 62 | 67 | 72 |
For each column, we'll multiply the frequency value f with its corresponding midpoint M
Eg: the first column has 7*42 = 294
| Marks | 40 - 44 | 45 - 49 | 50 - 54 | 55 - 59 | 60 - 64 | 65 - 69 | 70 - 74 | | No. of students (f) | 7 | 10 | 22 | 15 | 12 | 6 | 3 | | Midpoint (M) | 42 | 47 | 52 | 57 | 62 | 67 | 72 | | f*M | 294 | 470 | 1144 | 855 | 744 | 402 | 216 |
Adding up the values in that bottom row gets us
294 + 470 + 1144 + 855 + 744 + 402 + 216 = 4125
Divide that over the total number of students (75)
4125/75 = 55
The mean is 55.
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There are an odd number of students (75)
75/2 = 37.5 this rounds to 38
The median is in slot 38. The first 37 items are below it, the upper 37 items are above the median.
That gives 37+1+37 = 75 scores total.
Clearly the median is not in class 40-44 since we only have 7 students here.
The median is also not in the 45-49 class because 7+10 = 17 is too small also
7+10+22 = 39 is now too large since we want to land on 38.
The median mark is somewhere between 50 and 54.
We don't have enough info to determine its exact value because of the grouped nature of the marks.
The mode is the most frequent score.
The largest frequency is 22 and it occurs for the 50-54 class.
Like with the median, we can't pinpoint its exact value.
It's possible that the median = mode, but it's not a guarantee.
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Answers:
Mean = 55
Median = somewhere between 50 and 54.
Mode = somewhere between 50 and 54.
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