Statistics — Important Questions

1. A survey of 40 students regarding their performance in Mathematics was recorded as follows. Find the mean number of students who scored less than 30 marks.
Class interval Number of students
0 - 10 5
10 - 20 12
20 - 30 8
30 - 40 10
40 - 50 5
[3 marks]
2. The mean of the following frequency distribution is 62. Find the value of the missing frequency 'f'.
Class interval 0 - 20 20 - 40 40 - 60 60 - 80 80 - 100
Frequency 5 f 10 8 2
[3 marks]
3. The following table shows the distribution of students of two sections A and B according to the marks obtained by them.
Marks obtained | Section A | Section B

Answers

2. We will use the direct method to find the mean.
Class interval | Frequency ($f_i$) | Class Mark ($x_i$) | $f_i{x}_i$

| ----------------- | ----------------- | | ---------- | 0 - 20 | 5 | 10 | 50
20 - 40 | $f$ | 30 | $30f$
40 - 60 | 10 | 50 | 500
60 - 80 | 8 | 70 | 560
80 - 100 | 2 | 90 | 180
Total | $25+f$ | | $1290+30f$

Mean ($\bar{x}$) = $\frac{\sum f_i{x}_i}{\sum f_i}$

This calculation leads to a negative frequency, which is not possible. Let's recheck the question or the provided NCERT text for potential issues. Assuming there might be a typo and proceeding with the formula. If the mean was intended to be different, or the data points.

Let's assume the mean is correct and re-evaluate if there's an error in calculation or premise.
If the mean is 62, and the total frequency is sum of frequencies, let's assume $\sum f_i=N$.
If $f=5$, total frequency = 30, $\sum f_i{x}_i=1290+30(5)=1290+150=1440$. Mean = $1440\mathrm{/}30=48$.
If $f=10$, total frequency = 35, $\sum f_i{x}_i=1290+30(10)=1590$. Mean = $1590\mathrm{/}35=45.43$.

There seems to be an inconsistency in the problem statement as provided, leading to a negative frequency. Assuming there was a typo in the mean or the data, and illustrating the method:
Let's re-attempt with a hypothetical case where the calculation might be valid.
Suppose the mean was 40.

$f=29$. This would be a valid frequency.

Given the inconsistency, we cannot provide a numerical answer for $f$. However, the method is as shown above.
Answer: Method shown above. Calculation leads to an invalid result, indicating a potential issue with the question data.

(ii) To find the median marks for Section B, we first calculate the cumulative frequencies.
Class interval | Frequency ($f_i$) | Cumulative Frequency (cf)

Total frequency ($n$) = 70.
The median class is the class with the cumulative frequency greater than $n\mathrm{/}2=70\mathrm{/}2=35$.
The cumulative frequency 45 corresponds to the class 45 - 60. So, the median class is 45 - 60.
$l$ (lower limit of median class) = 45
$h$ (class size) = 15
$cf$ (cumulative frequency of class preceding the median class) = 33
$f$ (frequency of median class) = 12

Median = $l+\left(\frac{n\mathrm{/}2-cf}{f}\right)\times h$
Median = $45+\left(\frac{35-33}{12}\right)\times 15$
Median = $45+\left(\frac{2}{12}\right)\times 15$
Median = $45+\frac{1}{6}\times 15$
Median = $45+\frac{15}{6}$
Median = $45+2.5$
Median = 47.5
Answer: 47.5 marks.

4. To find the median age, we need to convert the 'less than' data into class intervals and then calculate the cumulative frequencies.
Class interval | Number of persons ($f_i$) | Cumulative Frequency (cf)

Total frequency ($n$) = 51.
$n\mathrm{/}2=51\mathrm{/}2=25.5$.
The median class is the class with the cumulative frequency just greater than 25.5, which is 29, corresponding to the class 145 - 150.
$l$ (lower limit of median class) = 145
$h$ (class size) = 5
$cf$ (cumulative frequency of class preceding the median class) = 11
$f$ (frequency of median class) = 18

Median = $l+\left(\frac{n\mathrm{/}2-cf}{f}\right)\times h$
Median = $145+\left(\frac{25.5-11}{18}\right)\times 5$
Median = $145+\left(\frac{14.5}{18}\right)\times 5$
Median = $145+\frac{72.5}{18}$
Median = $145+\mathrm{4.027...}$
Median $\approx 149.03$ (rounding to two decimal places as in the NCERT example)
Answer: 149.03 years.

6. We first calculate the cumulative frequencies and set up the equation for the median.
Class interval | Frequency ($f_i$) | Cumulative Frequency (cf)

Total frequency ($n$) = $35+p+q$. Given $n=60$.
So, $35+p+q=60\Rightarrow p+q=25$ (Equation 1).

The median is 52.5, which lies in the class interval 50 - 60.
$l$ (lower limit of median class) = 50
$h$ (class size) = 10
$cf$ (cumulative frequency of class preceding the median class) = $30+p$
$f$ (frequency of median class) = 5

Median = $l+\left(\frac{n\mathrm{/}2-cf}{f}\right)\times h$

$p=-2.5\mathrm{/}2=-1.25$.
A negative frequency is not possible. This indicates an issue with the problem data or the given median value. Let's re-examine the median class.
If the median is 52.5, it must fall in the 50-60 class.

Let's check if the median class could be 40-50. If median was 45, it would be in 40-50.
The median is 52.5, which is greater than 50. So the median class must be 50-60.

There seems to be an issue with the provided median value or the frequencies given. Assuming the formula and method are to be demonstrated:
Let's proceed by assuming the median class is correct and see if $p$ and $q$ can be found.

$2.5=-2p\Rightarrow p=-1.25$ which is impossible.

Let's consider the possibility that the median is actually in a different class, meaning there might be an error in the problem statement. However, if we must follow the given median of 52.5, and the class intervals, the median class is 50-60.

Let's assume there's a typo in the total frequency or the median value.
If we assume $p=10$, then $q=15$ from $p+q=25$. Total frequency would be $35+10+15=60$.
Class interval | Frequency ($f_i$) | Cumulative Frequency (cf)

Let's reconsider the original calculation for $p$ and $q$ with the given median 52.5.

$p=-1.25$. This is not a valid frequency.

Given the impossibility of obtaining positive frequencies with the provided median and data, there is likely an error in the question. However, if forced to provide an answer based on the structure of such questions, it would involve solving for $p$ and $q$ using the two equations: $p+q=25$ and the median formula. Since $p$ is coming out negative, it's impossible to proceed.
Answer: The problem statement likely contains an error, as it leads to negative frequencies.

7. We will use the step-deviation method.
Class interval | Frequency ($f_i$) | Class Mark ($x_i$) | $d_i=x_i-a$ | $u_i=d_i\mathrm{/}h$ | $f_i{u}_i$

| ----------------- | ----------------- | ----------------- | --------------- | | ---------- | 0 - 150 | 8 | 75 | -250 | -2.5 | -20
150 - 300 | 15 | 225 | -100 | -1 | -15
300 - 450 | 25 | 375 | 0 | 0 | 0
450 - 600 | 32 | 525 | 150 | 1.5 | 48
600 - 750 | 20 | 675 | 400 | 4 | 80
750 - 900 | 10 | 825 | 550 | 5.5 | 55

The class size is not uniform ($h=150$ for the first two, then $150$, then $150$, then $150$, then $150$). However, the problem states "step-deviation method", implying we should try to use it. Let's assume a common class width for the purpose of step-deviation, or use the class marks directly. Let's re-evaluate class marks and $h$.
Class interval | Frequency ($f_i$) | Class Mark ($x_i$) | Assume $a=375$ ($h=150$) | $u_i=(x_i-a)\mathrm{/}h$ | $f_i{u}_i$

| ----------------- | ----------------- | --------------------------- | ------------------- | | ---------- | 0 - 150 | 8 | 75 | $75-375=-300$ | $-300\mathrm{/}150=-2$ | -16
150 - 300 | 15 | 225 | $225-375=-150$ | $-150\mathrm{/}150=-1$ | -15
300 - 450 | 25 | 375 | $375-375=0$ | $0\mathrm{/}150=0$ | 0
450 - 600 | 32 | 525 | $525-375=150$ | $150\mathrm{/}150=1$ | 32
600 - 750 | 20 | 675 | $675-375=300$ | $300\mathrm{/}150=2$ | 40
750 - 900 | 10 | 825 | $825-375=450$ | $450\mathrm{/}150=3$ | 30
Total | $\sum f_i=110$ | | | | $\sum f_i{u}_i=111$

Mean ($\bar{x}$) = $a+\left(\frac{\sum f_i{u}_i}{\sum f_i}\right)\times h$
Mean ($\bar{x}$) = $375+\left(\frac{111}{110}\right)\times 150$
Mean ($\bar{x}$) = $375+\mathrm{1.00909...}\times 150$
Mean ($\bar{x}$) = $375+\mathrm{151.3636...}$
Mean ($\bar{x}$) $\approx 526.36$

Let's recheck the problem statement and NCERT examples for step-deviation with unequal class sizes. NCERT Example 3 suggests it's possible, but may require careful handling.

Let's try using the direct method as it's more robust for unequal class sizes if step-deviation is problematic.
Class interval | Frequency ($f_i$) | Class Mark ($x_i$) | $f_i{x}_i$

| ----------------- | ----------------- | | ---------- | 0 - 150 | 8 | 75 | 600
150 - 300 | 15 | 225 | 3375
300 - 450 | 25 | 375 | 9375
450 - 600 | 32 | 525 | 16800
600 - 750 | 20 | 675 | 13500
750 - 900 | 10 | 825 | 8250
Total | $\sum f_i=110$ | | $\sum f_i{x}_i=51900$

Mean ($\bar{x}$) = $\frac{\sum f_i{x}_i}{\sum f_i}=\frac{51900}{110}=\frac{5190}{11}=\mathrm{471.8181...}$
Mean ($\bar{x}$) $\approx 471.82$

The question specifically asks for the step-deviation method. Let's re-verify the class marks and $h$ for step-deviation.
The class intervals are 0-150, 150-300, 300-450, 450-600, 600-750, 750-900. The class width is 150 for all.
Let's use $a=375$ (midpoint of 300-450 and 450-600) and $h=150$.
Class interval | Frequency ($f_i$) | Class Mark ($x_i$) | $d_i=x_i-375$ | $u_i=d_i\mathrm{/}150$ | $f_i{u}_i$

| ----------------- | ----------------- | ----------------- | --------------- | | ---------- | 0 - 150 | 8 | 75 | $75-375=-300$ | $-300\mathrm{/}150=-2$ | -16
150 - 300 | 15 | 225 | $225-375=-150$ | $-150\mathrm{/}150=-1$ | -15
300 - 450 | 25 | 375 | $375-375=0$ | $0\mathrm{/}150=0$ | 0
450 - 600 | 32 | 525 | $525-375=150$ | $150\mathrm{/}150=1$ | 32
600 - 750 | 20 | 675 | $675-375=300$ | $300\mathrm{/}150=2$ | 40
750 - 900 | 10 | 825 | $825-375=450$ | $450\mathrm{/}150=3$ | 30
Total | $\sum f_i=110$ | | | | $\sum f_i{u}_i=71$

Mean ($\bar{x}$) = $a+\left(\frac{\sum f_i{u}_i}{\sum f_i}\right)\times h$
Mean ($\bar{x}$) = $375+\left(\frac{71}{110}\right)\times 150$
Mean ($\bar{x}$) = $375+\mathrm{0.64545...}\times 150$
Mean ($\bar{x}$) = $375+\mathrm{96.8181...}$
Mean ($\bar{x}$) $\approx 471.82$
Answer: Rs 471.82.

Given the exact wording "mean of five numbers is 15", and "mean of the remaining 9 observations is 16", the question is contradictory. Assuming a typo, and that it should have been 10 observations initially:
Sum of 10 observations = $15\times 10=150$.
Sum of remaining 9 observations = $16\times 9=144$.
Removed observation = $150-144=6$.
Answer: 6.

9. To find the median rainfall, we construct the cumulative frequency table.
Class interval (Rainfall in mm) | Number of days ($f_i$) | Cumulative Frequency (cf)

Total frequency ($n$) = 50.
$n\mathrm{/}2=50\mathrm{/}2=25$.
The median class is the class with the cumulative frequency just greater than 25, which is 32, corresponding to the class 20 - 30.
$l$ (lower limit of median class) = 20
$h$ (class size) = 10
$cf$ (cumulative frequency of class preceding the median class) = 17
$f$ (frequency of median class) = 15

Median = $l+\left(\frac{n\mathrm{/}2-cf}{f}\right)\times h$
Median = $20+\left(\frac{25-17}{15}\right)\times 10$
Median = $20+\left(\frac{8}{15}\right)\times 10$
Median = $20+\frac{80}{15}$
Median = $20+\frac{16}{3}$
Median = $20+\mathrm{5.333...}$
Median $\approx 25.33$ mm.
Answer: 25.33 mm.

Mode = $l+\left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right)\times h$

$f=46\mathrm{/}4=11.5$.
A frequency cannot be a decimal. This suggests an error in the problem statement or given mode.

Let's re-evaluate the interpretation of $f_0$ and $f_2$.
Frequency: 10, 8, $f$, 12, 10.
Mode = 32. This falls in 30-40.
Modal class = 30-40.
$l=30$, $h=10$, $f_1=12$.
$f_0$ is the frequency of the class preceding the modal class. This is the class 20-30, so $f_0=f$.
$f_2$ is the frequency of the class succeeding the modal class. This is the class 40-50, so $f_2=10$.

Mode = $l+\left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right)\times h$

$f=9.2\mathrm{/}0.8=92\mathrm{/}8=11.5$.

Still a decimal frequency. Let's assume the modal class is 20-30 if $f$ is greater than 12.
If modal class is 20-30: $l=20,h=10,f_1=f,f_0=8,f_2=12$.

$f=16\mathrm{/}1.4=160\mathrm{/}14=80\mathrm{/}7\approx 11.43$. Still decimal.

There is a high probability of an error in the question. Assuming the first calculation (modal class 30-40) is the intended one, and a decimal frequency is an artifact of the problem setter.
Answer: 11.5. (Note: This is an invalid frequency).

Mode = $l+\left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right)\times h$
Mode = $3+\left(\frac{8-7}{2(8)-7-2}\right)\times 2$
Mode = $3+\left(\frac{1}{16-7-2}\right)\times 2$
Mode = $3+\left(\frac{1}{7}\right)\times 2$
Mode = $3+\frac{2}{7}$
Mode = $3+\mathrm{0.2857...}$
Mode $\approx 3.286$.
Answer: 3.286.

12. To find the mean marks, we use the direct method.
Class interval | Frequency ($f_i$) | Class Mark ($x_i$) | $f_i{x}_i$

| ----------------- | ----------------- | | ---------- | 0 - 10 | 5 | 5 | 25
10 - 20 | 15 | 15 | 225
20 - 30 | 25 | 25 | 625
30 - 40 | 30 | 35 | 1050
40 - 50 | 25 | 45 | 1125
Total | $\sum f_i=100$ | | $\sum f_i{x}_i=3050$

Mean ($\bar{x}$) = $\frac{\sum f_i{x}_i}{\sum f_i}=\frac{3050}{100}=30.5$
Answer: 30.5 marks.

13. To find the mean percentage of female teachers, we use the direct method.
Class interval | Frequency ($f_i$) | Class Mark ($x_i$) | $f_i{x}_i$

| ----------------- | ----------------- | | ---------- | 15 - 25 | 6 | 20 | 120
25 - 35 | 11 | 30 | 330
35 - 45 | 7 | 40 | 280
45 - 55 | 4 | 50 | 200
55 - 65 | 4 | 60 | 240
65 - 75 | 2 | 70 | 140
75 - 85 | 1 | 80 | 80
Total | $\sum f_i=35$ | | $\sum f_i{x}_i=1390$

Mean ($\bar{x}$) = $\frac{\sum f_i{x}_i}{\sum f_i}=\frac{1390}{35}=\frac{278}{7}=\mathrm{39.714...}$
Answer: 39.71%.

$x=10$.
The first three observations are $x$, $x+2$, $x+4$.
These are $10$, $10+2=12$, $10+4=14$.
The mean of the first three observations is $\frac{10+12+14}{3}=\frac{36}{3}=12$.
Answer: 12.

15. To find the mean number of days a student was absent, we use the direct method.
Class interval (Number of days absent) | Number of students ($f_i$) | Class Mark ($x_i$) | $f_i{x}_i$

| -------------------------- | ----------------- | | ---------- | 0 - 6 | 11 | 3 | 33
6 - 10 | 10 | 8 | 80
10 - 14 | 7 | 12 | 84
14 - 20 | 4 | 17 | 68
20 - 28 | 4 | 24 | 96
28 - 38 | 3 | 33 | 99
38 - 40 | 1 | 39 | 39
Total | $\sum f_i=40$ | | $\sum f_i{x}_i=399$

Mean ($\bar{x}$) = $\frac{\sum f_i{x}_i}{\sum f_i}=\frac{399}{40}=9.975$
Answer: 9.975 days.

16. We first calculate the cumulative frequencies.
Class interval | Frequency ($f_i$) | Cumulative Frequency (cf)

Total frequency ($n$) = $x+q+47$. Given $n=100$.
So, $x+q+47=100\Rightarrow x+q=53$ (Equation 1).

The median is 52.5, which lies in the class interval 50 - 60.
$l$ (lower limit of median class) = 50
$h$ (class size) = 10
$cf$ (cumulative frequency of class preceding the median class) = $x+q+27$
$f$ (frequency of median class) = 8

Median = $l+\left(\frac{n\mathrm{/}2-cf}{f}\right)\times h$

Substitute $x+q=53$ from Equation 1:

$2.5=-37.5$.
This is a contradiction. There must be an error in the problem statement (median value or total frequency or given frequencies).

Let's assume the median is in the 40-50 class and re-solve. This is not mathematically sound, but to demonstrate the process if the problem were correct.
If median class was 40-50: $l=40,h=10,cf=x+27,f=q$.

$2.25q=23-x$.
We have $x+q=53$, so $x=53-q$.

$q=-30\mathrm{/}1.25=-24$. This is also impossible.

Given the inconsistency, we cannot provide a valid numerical answer for $p$ and $q$.
Answer: The problem statement likely contains an error, as it leads to contradictions.

17. To find the mean concentration, we use the direct method.
Class interval (Concentration of $S{O}_2$ in ppm) | Frequency ($f_i$) | Class Mark ($x_i$) | $f_i{x}_i$

| ----------------- | ----------------- | | ---------- | 0.00 - 0.04 | 15 | 0.02 | 0.30
0.04 - 0.08 | 25 | 0.06 | 1.50
0.08 - 0.12 | 30 | 0.10 | 3.00
0.12 - 0.16 | 18 | 0.14 | 2.52
0.16 - 0.20 | 12 | 0.18 | 2.16
0.20 - 0.24 | 2 | 0.22 | 0.44
Total | $\sum f_i=102$| | $\sum f_i{x}_i=9.92$

Mean ($\bar{x}$) = $\frac{\sum f_i{x}_i}{\sum f_i}=\frac{9.92}{102}=\mathrm{0.09725...}$
Answer: 0.097 ppm (approximately).

18. To find the missing frequency $f$, we use the formula for the mean.
Daily pocket allowance (in `) | Number of children ($f_i$) | Class Mark ($x_i$) | $f_i{x}_i$

| -------------------------- | ----------------- | | ---------- | 11 - 13 | 7 | 12 | 84
13 - 15 | 6 | 14 | 84
15 - 17 | 9 | 16 | 144
17 - 19 | 13 | 18 | 234
19 - 21 | $f$ | 20 | $20f$
21 - 23 | 5 | 22 | 110
23 - 25 | 4 | 24 | 96
Total | $\sum f_i=44+f$ | | $\sum f_i{x}_i=752+20f$

Mean ($\bar{x}$) = 18.

$f=20$.
Answer: 20.

19. To find the median number of runs scored, we need to calculate cumulative frequencies.
Runs scored | Number of batsmen ($f_i$) | Cumulative Frequency (cf)

Median = $l+\left(\frac{n\mathrm{/}2-cf}{f}\right)\times h$
Median = $5000+\left(\frac{24.5-22}{9}\right)\times 1000$
Median = $5000+\left(\frac{2.5}{9}\right)\times 1000$
Median = $5000+\frac{2500}{9}$
Median = $5000+\mathrm{277.777...}$
Median $\approx 5277.78$.
Answer: Rs 5277.78.

$x=19\mathrm{/}3$.

Now we need to find the median of the data. First, substitute the value of $x$ into the observations.
The observations are:
2
9

5
10
5

The observations are: 2, 9, 37/3, 47/3, 5, 10, 5.
Let's write them in ascending order:
2, 5, 5, 9, 10, 37/3, 47/3.
(Approximate values: 2, 5, 5, 9, 10, 12.33, 15.67)
There are 7 observations. The median is the middle observation (4th observation).
The 4th observation is 9.
Answer: 9.