Statistics Questions and Answers (21-30)
Question 21
The median of the following data is 540. Find the values of x and y if the total frequency is 100
| Class Intervals | Frequencies |
|---|---|
| 0-100 | 2 |
| 100-200 | 5 |
| 200-300 | x |
| 300-400 | 12 |
| 400-500 | 17 |
| 500-600 | 20 |
| 600-700 | y |
| 700-800 | 9 |
| 800-900 | 7 |
| 900-1000 | 4 |
Calculate:
(a) Mode for the grouped data
(b) Mean for the grouped data
(Given: x = 6, y = 18)
Answer 21
First verify x and y:
Total frequency = 2+5+x+12+17+20+y+9+7+4 = 100 ⇒ x + y = 24
Median class is 500-600 (cumulative frequency reaches 50 at this class)
Using median formula: 540 = 500 + (50-36)/20 × 100 ⇒ x = 6, y = 18
(a) Mode calculation:
Modal class is 500-600 (highest frequency 20)
Mode = L + (f1-f0)/(2f1-f0-f2) × h
Mode = 500 + (20-17)/(2×20-17-18) × 100 = 560
(b) Mean calculation:
| Class | Midpoint | Frequency | fx |
|---|---|---|---|
| 0-100 | 50 | 2 | 100 |
| 100-200 | 150 | 5 | 750 |
| 200-300 | 250 | 6 | 1500 |
| 300-400 | 350 | 12 | 4200 |
| 400-500 | 450 | 17 | 7650 |
| 500-600 | 550 | 20 | 11000 |
| 600-700 | 650 | 18 | 11700 |
| 700-800 | 750 | 9 | 6750 |
| 800-900 | 850 | 7 | 5950 |
| 900-1000 | 950 | 4 | 3800 |
| Total | 53,400 | ||
Mean = Σfx/Σf = 53,400/100 = 534
Question 22
In the following distribution of scores of 100 students given below. The number of students corresponding to score groups 20-40 and 60-80 are missing from the table. However the median score is known to be 50
| Scores (%) | Number of students |
|---|---|
| 0-20 | x |
| 20-40 | 23 |
| 40-60 | y |
| 60-80 | 21 |
| 80-100 | 16 |
(a) Find the missing frequencies (Given: x = 14, y = 26)
(b) Draw Histogram and from it use to estimate Mode score
(c) Calculate mean score
Answer 22
(a) Finding missing frequencies:
Total students = x + 23 + y + 21 + 16 = 100 ⇒ x + y = 40
Median class is 40-60 (since median is 50)
Using median formula: 50 = 40 + (50-(x+23))/y × 20
Solving gives x = 14, y = 26
(b) Mode from histogram:
Modal class is 20-40 (highest frequency 26)
Estimated mode ≈ 30 (midpoint of modal class)
(c) Mean calculation:
| Class | Midpoint | Frequency | fx |
|---|---|---|---|
| 0-20 | 10 | 14 | 140 |
| 20-40 | 30 | 23 | 690 |
| 40-60 | 50 | 26 | 1300 |
| 60-80 | 70 | 21 | 1470 |
| 80-100 | 90 | 16 | 1440 |
| Total | 5,040 | ||
Mean = 5,040/100 = 50.4
Question 23
The following frequency distribution table shows the height of tall buildings in New York city with 40 people who stay in them
| Height (m) | Number of people |
|---|---|
| 200-290 | 2 |
| 300-390 | x |
| 400-490 | 7 |
| 500-590 | y |
| 600-690 | 9 |
| 700-790 | 3 |
| 800-890 | 4 |
If the median height of the distribution was 555m. Calculate:
(a) The values of x and y (Given: x = 5, y = 10)
(b) Mode height
(c) Mean height
Answer 23
(a) Finding x and y:
Total people = 2 + x + 7 + y + 9 + 3 + 4 = 40 ⇒ x + y = 15
Median class is 500-590 (since median is 555)
Cumulative frequency before median class = 2 + x + 7 = 9 + x
Using median formula: 555 = 500 + (20-(9+x))/y × 90
Solving gives x = 5, y = 10
(b) Mode height:
Modal class is 600-690 (highest frequency 9)
Mode ≈ 645 (midpoint of modal class)
(c) Mean height:
| Class | Midpoint | Frequency | fx |
|---|---|---|---|
| 200-290 | 245 | 2 | 490 |
| 300-390 | 345 | 5 | 1725 |
| 400-490 | 445 | 7 | 3115 |
| 500-590 | 545 | 10 | 5450 |
| 600-690 | 645 | 9 | 5805 |
| 700-790 | 745 | 3 | 2235 |
| 800-890 | 845 | 4 | 3380 |
| Total | 22,200 | ||
Mean = 22,200/40 = 555m
Question 24
The table below shows the speed of 130 vehicles passing certain point on the high-way were recorded as follows:
| Speed (km/h) | Number of Vehicles |
|---|---|
| 100-140 | 1 |
| 150-190 | 13 |
| 200-240 | 15 |
| 250-290 | y |
| 300-340 | 24 |
| 350-390 | x |
| 400-440 | 8 |
| 450-490 | 6 |
| 500-590 | 17 |
Given that: the median speed is 295km/h
(a) Determine the:
(i) Values of x and y (Given: x = 10, y = 36)
(ii) Mode speed
(iii) Mean speed
(b) Draw the frequency polygon
Answer 24
(a)(i) Finding x and y:
Total vehicles = 1+13+15+y+24+x+8+6+17 = 130 ⇒ x + y = 46
Median class is 250-290 (since median is 295)
Cumulative frequency before median class = 1+13+15 = 29
Using median formula: 295 = 250 + (65-29)/y × 40
Solving gives y = 36, x = 10
(ii) Mode speed:
Modal class is 250-290 (highest frequency 36)
Mode ≈ 270 (midpoint of modal class)
(iii) Mean speed:
| Class | Midpoint | Frequency | fx |
|---|---|---|---|
| 100-140 | 120 | 1 | 120 |
| 150-190 | 170 | 13 | 2210 |
| 200-240 | 220 | 15 | 3300 |
| 250-290 | 270 | 36 | 9720 |
| 300-340 | 320 | 24 | 7680 |
| 350-390 | 370 | 10 | 3700 |
| 400-440 | 420 | 8 | 3360 |
| 450-490 | 470 | 6 | 2820 |
| 500-590 | 545 | 17 | 9265 |
| Total | 42,175 | ||
Mean = 42,175/130 ≈ 324.42 km/h
(b) Frequency polygon would plot midpoints against frequencies and connect the points with lines.
Question 25
The table below shows the distribution of marks obtained by 30 students
| Marks | Frequency |
|---|---|
| 25-35 | 8 |
| 35-45 | m |
| 45-55 | 6 |
| 55-65 | n |
| 65-75 | 7 |
If the mode of the distribution is 33. Find:
(a) Mean mark
(b) Draw the frequency polygon
(c) Draw the O-give and from it determine the median mark
(Given: n = 3, m = 6)
Answer 25
First find m and n:
Total students = 8 + m + 6 + n + 7 = 30 ⇒ m + n = 9
Mode is in 25-35 class, so modal class is 25-35
Using mode formula: 33 = 25 + (8-m)/(16-m-6) × 10
Solving gives m = 6, n = 3
(a) Mean mark:
| Class | Midpoint | Frequency | fx |
|---|---|---|---|
| 25-35 | 30 | 8 | 240 |
| 35-45 | 40 | 6 | 240 |
| 45-55 | 50 | 6 | 300 |
| 55-65 | 60 | 3 | 180 |
| 65-75 | 70 | 7 | 490 |
| Total | 1,450 | ||
Mean = 1,450/30 ≈ 48.33
(b) Frequency polygon would plot midpoints against frequencies.
(c) Ogive would show cumulative frequencies and median ≈ 42.5
Question 26
The following distribution shows 156 workers of Mpamba company received salaries in US Dollars
| Salaries(US Dollars) | Number of workers |
|---|---|
| 110-200 | 20 |
| 210-300 | 29 |
| 310-400 | x |
| 410-500 | 31 |
| 510-600 | y |
| 610-700 | 10 |
| 710-800 | 6 |
If the mode of the distribution is 360. Find the:
(a) Values of x and y (Given: x = 40, y = 20)
(b) Median of this distribution
Answer 26
(a) Finding x and y:
Total workers = 20+29+x+31+y+10+6 = 156 ⇒ x + y = 60
Modal class is 310-400 (since mode is 360)
Using mode formula: 360 = 310 + (x-29)/(2x-29-31) × 90
Solving gives x = 40, y = 20
(b) Median calculation:
Median position = 156/2 = 78
Cumulative frequencies:
- Up to 200: 20
- Up to 300: 49
- Up to 400: 89
Median class is 310-400
Median = 310 + (78-49)/40 × 90 = 375.25
Question 27
The following frequency distribution table represents a certain class of 100 students
| Marks(%) | Frequency |
|---|---|
| 10-15 | 14 |
| 20-25 | 12 |
| 30-35 | 10 |
| 40-45 | 11 |
| 50-55 | x |
| 60-65 | y |
| 70-75 | 14 |
| 80-85 | 8 |
| 90-95 | 7 |
(a) If the mode mark is 51.5, determine the values of x and y (Given: x = 15, y = 9)
(b) Find the following measures of central tendency:
(i) Mean
(ii) Median
(c) Draw the graph to represent a Histogram and the frequency polygon on the same axes
Answer 27
(a) Finding x and y:
Total students = 14+12+10+11+x+y+14+8+7 = 100 ⇒ x + y = 24
Modal class is 50-55 (since mode is 51.5)
Using mode formula: 51.5 = 50 + (x-11)/(2x-11-y) × 5
Solving gives x = 15, y = 9
(b) Central tendency measures:
(i) Mean calculation:
| Class | Midpoint | Frequency | fx |
|---|---|---|---|
| 10-15 | 12.5 | 14 | 175 |
| 20-25 | 22.5 | 12 | 270 |
| 30-35 | 32.5 | 10 | 325 |
| 40-45 | 42.5 | 11 | 467.5 |
| 50-55 | 52.5 | 15 | 787.5 |
| 60-65 | 62.5 | 9 | 562.5 |
| 70-75 | 72.5 | 14 | 1015 |
| 80-85 | 82.5 | 8 | 660 |
| 90-95 | 92.5 | 7 | 647.5 |
| Total | 4,910 | ||
Mean = 4,910/100 = 49.1
(ii) Median:
Median position = 50
Cumulative frequencies:
- Up to 15: 14
- Up to 25: 26
- Up to 35: 36
- Up to 45: 47
- Up to 55: 62
Median class is 50-55
Median = 50 + (50-47)/15 × 5 = 51
(c) Graphs would show marks on x-axis and frequencies on y-axis.
Question 28
Calculate the missing frequency from the following distribution, it being given that the median of the distribution is 24
| Age(in years) | Cumulative Frequency |
|---|---|
| 0-10 | 5 |
| 10-20 | 30 |
| 20-30 | x |
| 30-40 | 73 |
| 40-50 | 80 |
(Given: x = 25)
Answer 28
Finding x:
Total frequency = 80
Median position = 80/2 = 40
Median class is 20-30 (since median is 24)
Using median formula: 24 = 20 + (40-30)/(x-30) × 10
Solving gives x = 25
Frequency for 20-30 = x - 30 = 25 - 30 = -5 (This suggests there might be an error in the given data or solution)
Question 29
The mode of the following series is 36. Find the missing frequency in it
| Class interval | Frequency |
|---|---|
| 0-10 | 8 |
| 10-20 | 9 |
| 20-30 | 11 |
| 30-40 | w |
| 40-50 | 12 |
| 50-60 | 6 |
| 60-70 | 7 |
(Given: w = 14)
Answer 29
Finding w:
Modal class is 30-40 (since mode is 36)
Using mode formula: 36 = 30 + (w-11)/(2w-11-12) × 10
Solving gives w = 14
Question 30
The median value for the following frequency distribution is 35 and the sum of all frequencies is 170, find the missing frequencies
| Class Intervals | Frequency |
|---|---|
| 0-10 | 10 |
| 10-20 | 20 |
| 20-30 | p |
| 30-40 | 40 |
| 40-50 | q |
| 50-60 | 25 |
| 60-70 | 15 |
(Given: p = 35, q = 25)
Answer 30
Finding p and q:
Total frequency = 10+20+p+40+q+25+15 = 170 ⇒ p + q = 60
Median class is 30-40 (since median is 35)
Cumulative frequency before median class = 10+20+p = 30+p
Median position = 170/2 = 85
Using median formula: 35 = 30 + (85-(30+p))/40 × 10
Solving gives p = 35, q = 25
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