Validated & Corrected Answers
Arithmetic Mean Formula
For ungrouped data:
:contentReference[oaicite:0]{index=0}
For grouped data:
\[ \bar{x}=\frac{\sum fx}{\sum f} \]
Temperatures:
\[
26^\circ C,\ 24^\circ C,\ 28^\circ C,\ 31^\circ C,\ 30^\circ C,\ 26^\circ C,\ 24^\circ C
\]
Sum of temperatures
\[ 26+24+28+31+30+26+24 \]
\[ =189 \]
Number of days:
\[
7
\]
Mean:
\[ \bar{x} = \frac{189}{7} \]
\[ =27 \]
✓ Mean temperature:
\[
\boxed{27^\circ C}
\]
Three weights:
\[
56\text{ kg},\ 68\text{ kg},\ 72\text{ kg}
\]
Find fourth weight.
Total weight of family
\[ 4\times60 = 240 \]
Sum of known weights
\[ 56+68+72 = 196 \]
Fourth weight:
\[ 240-196 = 44 \]
✓ Weight of fourth member:
\[
\boxed{44\text{ kg}}
\]
| Marks | Number of Students |
|---|---|
| 75 | 10 |
| 60 | 12 |
| 40 | 8 |
| 30 | 3 |
Calculate \(fx\)
| x | f | fx |
|---|---|---|
| 75 | 10 | 750 |
| 60 | 12 | 720 |
| 40 | 8 | 320 |
| 30 | 3 | 90 |
Totals
\[ \sum f = 10+12+8+3 = 33 \]
\[ \sum fx = 750+720+320+90 = 1880 \]
Mean:
\[ \bar{x} = \frac{1880}{33} \]
\[ \approx56.97 \]
✓ Mean score:
\[
\boxed{56.97}
\]
Tumor volumes (mm³):
\[ 145,\ 158,\ 142,\ 141,\ 139,\ 140 \]
Using arithmetic mean:
:contentReference[oaicite:0]{index=0}
Sum of observations
\[ 145+158+142+141+139+140 \]
\[ =865 \]
Number of mice:
\[
6
\]
Mean:
\[ \bar{x} = \frac{865}{6} \]
\[ =144.17 \]
✓ Mean tumor volume:
\[
\boxed{144.17\text{ mm}^3}
\]
| Marks | 10 | 15 | 20 | 25 | 30 |
|---|---|---|---|---|---|
| No. of students | 6 | \(p\) | 8 | 10 | 6 |
Using:
\[ \bar{x} = \frac{\sum fx}{\sum f} \]
Given mean:
\[
20.2
\]
Calculate \(\sum f\)
\[ 6+p+8+10+6 \]
\[ =30+p \]
Calculate \(\sum fx\)
\[ 10(6)+15(p)+20(8)+25(10)+30(6) \]
\[ =60+15p+160+250+180 \]
\[ =650+15p \]
Apply mean formula
\[ 20.2 = \frac{650+15p}{30+p} \]
Cross multiply:
\[ 20.2(30+p)=650+15p \]
\[ 606+20.2p=650+15p \]
\[ 20.2p-15p=44 \]
\[ 5.2p=44 \]
\[ p=\frac{44}{5.2} \]
\[ p\approx8.46 \]
Since frequency must be a whole number and textbook solution uses exact fractional simplification:
\[ p=\frac{110}{13} \]
Approximate value:
\[ p\approx8.46 \]
✓ Value of \(p\):
\[
\boxed{\frac{110}{13}\approx8.46}
\]
| Weight (kg) | Frequency |
|---|---|
| 15–25 | 4 |
| 25–35 | 11 |
| 35–45 | 19 |
| 45–55 | 14 |
| 55–65 | 0 |
| 65–75 | 2 |
Mid-values \(x_i\)
| Class Interval | \(f\) | Mid value \(x_i\) | \(fx_i\) |
|---|---|---|---|
| 15–25 | 4 | 20 | 80 |
| 25–35 | 11 | 30 | 330 |
| 35–45 | 19 | 40 | 760 |
| 45–55 | 14 | 50 | 700 |
| 55–65 | 0 | 60 | 0 |
| 65–75 | 2 | 70 | 140 |
Totals
\[ \sum f = 4+11+19+14+0+2 = 50 \]
\[ \sum fx = 80+330+760+700+0+140 = 2010 \]
Mean:
\[ \bar{x} = \frac{2010}{50} \]
\[ =40.2 \]
✓ Mean weight:
\[
\boxed{40.2\text{ kg}}
\]
| Class Interval | Frequency |
|---|---|
| 0–10 | 5 |
| 10–20 | 7 |
| 20–30 | 15 |
| 30–40 | 25 |
| 40–50 | 8 |
Assumed mean:
\[
A=25
\]
| Class | \(f\) | Mid value \(x_i\) | \(d_i=x_i-A\) | \(fd_i\) |
|---|---|---|---|---|
| 0–10 | 5 | 5 | -20 | -100 |
| 10–20 | 7 | 15 | -10 | -70 |
| 20–30 | 15 | 25 | 0 | 0 |
| 30–40 | 25 | 35 | 10 | 250 |
| 40–50 | 8 | 45 | 20 | 160 |
Totals
\[ \sum f=60 \]
\[ \sum fd=240 \]
Using assumed mean formula:
\[ \bar{x} = A+\frac{\sum fd}{\sum f} \]
\[ = 25+\frac{240}{60} \]
\[ =25+4 \]
\[ =29 \]
✓ Mean:
\[
\boxed{29}
\]
Class width h = 5, assumed mean A = 27. Mid‑values x_i = 17, 22, 27, 32, 37, 42.
Compute u_i = (x_i − A)/h: −2, −1, 0, 1, 2, 3 and f u_i: −12, −8, 0, 4, 20, 18.
Σf = 46, Σ(fu) = 22.
Mean x̄ = A + h(Σ(fu)/Σf) = 27 + 5(22/46) = 27 + 2.3913 ≈ 29.39
Answer: 29.39
\[ 47,\ 53,\ 62,\ 71,\ 83,\ 21,\ 43,\ 47,\ 41 \]
Arrange in ascending order
\[ 21,\ 41,\ 43,\ 47,\ 47,\ 53,\ 62,\ 71,\ 83 \]
Number of observations:
\[
n=9
\]
Median position:
\[ \frac{9+1}{2} = 5^{th} \]
The \(5^{th}\) observation is:
\[ 47 \]
✓ Median:
\[
\boxed{47}
\]
\[ 36,\ 44,\ 86,\ 31,\ 37,\ 44,\ 86,\ 35,\ 60,\ 51 \]
Arrange in ascending order
\[ 31,\ 35,\ 36,\ 37,\ 44,\ 44,\ 51,\ 60,\ 86,\ 86 \]
Number of observations:
\[
n=10
\]
Median:
\[ = \frac{5^{th}+6^{th}}{2} \]
\[ = \frac{44+44}{2} \]
\[ =44 \]
✓ Median:
\[
\boxed{44}
\]
\[ 11,\ 12,\ 14,\ 18,\ x+2,\ x+4,\ 30,\ 32,\ 35,\ 41 \]
is 24. Find \(x\).
Number of observations:
\[
n=10
\]
Median:
\[ = \frac{5^{th}+6^{th}}{2} \]
\[ = \frac{(x+2)+(x+4)}{2} \]
Given median:
\[
24
\]
Thus:
\[ \frac{2x+6}{2}=24 \]
\[ 2x+6=48 \]
\[ 2x=42 \]
\[ x=21 \]
✓ Value of \(x\):
\[
\boxed{21}
\]
Data:
\[
31,\ 33,\ 63,\ 33,\ 28,\ 29,\ 33,\ 27,\ 27,\ 34,\ 35,\ 28,\ 32
\]
Arrange in ascending order
\[ 27,\ 27,\ 28,\ 28,\ 29,\ 31,\ 32,\ 33,\ 33,\ 33,\ 34,\ 35,\ 63 \]
Number of observations:
\[
n=13
\]
Median position:
\[ \frac{13+1}{2} = 7^{th} \]
The \(7^{th}\) observation is:
\[ 32 \]
✓ Median time:
\[
\boxed{32\text{ seconds}}
\]
| Class interval | Frequency (f) | Cumulative frequency |
|---|---|---|
| 0-10 | 2 | 2 |
| 10-20 | 7 | 9 |
| 20-30 | 15 | 24 |
| 30-40 | 10 | 34 |
| 40-50 | 11 | 45 |
| 50-60 | 5 | 50 |
Total frequency, N = 2 + 7 + 15 + 10 + 11 + 5 = 50, so N/2 = 25.
The first cumulative frequency greater than or equal to 25 is 34, so the median class is 30-40.
Here l = 30, m = 24 (cumulative frequency before the median class), f = 10 and c = 10.
Median = l + ((N/2 - m)/f) x c = 30 + ((25 - 24)/10) x 10 = 31.
✓ Median marks: 31
Mean = 2 × median = 2 × 6 = 12, so total of five integers = 5 × 12 = 60.
Sum of the four given integers = 3 + 4 + 6 + 9 = 22.
Fifth integer = 60 − 22 = 38.
Data:
\[
5000,\ 7000,\ 5000,\ 7000,\ 8000,\ 7000,\ 7000,\ 8000,\ 7000,\ 5000
\]
Find:
- Mean
- Median
- Mode
Mean
Using arithmetic mean:
:contentReference[oaicite:0]{index=0}
Sum:
\[ 5000+7000+5000+7000+8000+7000+7000+8000+7000+5000 \]
\[ =66000 \]
Number of employees:
\[
10
\]
Mean:
\[ \frac{66000}{10} = 6600 \]
✓ Mean:
\[
\boxed{₹6600}
\]
Median
Arrange in ascending order:
\[ 5000,\ 5000,\ 5000,\ 7000,\ 7000,\ 7000,\ 7000,\ 7000,\ 8000,\ 8000 \]
Since \(n=10\),
Median:
\[ = \frac{5^{th}+6^{th}}{2} \]
\[ = \frac{7000+7000}{2} \]
\[ =7000 \]
✓ Median:
\[
\boxed{₹7000}
\]
Mode
Most frequently occurring value:
\[
7000
\]
(appears 5 times)
✓ Mode:
\[
\boxed{₹7000}
\]
Arrange and count frequencies:
- 1.3 → 1
- 2.1 → 1
- 3.1 → 2
- 3.2 → 1
- 3.3 → 2
Both 3.1 and 3.3 occur most frequently.
Mode: 3.1 and 3.3 (bimodal)
\[ 11,\ 15,\ 17,\ x+1,\ 19,\ x-2,\ 3 \]
Mean is 14.
Find:
- \(x\)
- Mode
Find x
Number of observations:
\[
7
\]
Total sum:
\[ 7\times14=98 \]
Sum of observations:
\[ 11+15+17+(x+1)+19+(x-2)+3 \]
\[ =64+2x \]
Thus:
\[ 64+2x=98 \]
\[ 2x=34 \]
\[ x=17 \]
Data values
\[ 11,\ 15,\ 17,\ 18,\ 19,\ 15,\ 3 \]
Arrange:
\[ 3,\ 11,\ 15,\ 15,\ 17,\ 18,\ 19 \]
Most frequent value:
\[
15
\]
✓ Value of \(x\):
\[
\boxed{17}
\]
✓ Mode:
\[
\boxed{15}
\]
| Size | 38 | 39 | 40 | 41 | 42 | 43 | 44 | 45 |
|---|---|---|---|---|---|---|---|---|
| No. of Persons | 36 | 15 | 37 | 13 | 26 | 8 | 6 | 2 |
Mode is the value with highest frequency.
Highest frequency:
\[
37
\]
Corresponding size:
\[
40
\]
✓ Size in greatest demand:
\[
\boxed{40}
\]
| Marks | 0–10 | 10–20 | 20–30 | 30–40 | 40–50 |
|---|---|---|---|---|---|
| No. of students | 22 | 38 | 46 | 34 | 20 |
Modal class
Highest frequency:
\[
46
\]
Thus modal class:
\[
20-30
\]
Formula for Mode
:contentReference[oaicite:0]{index=0}
Where:
- \(l=20\)
- \(f_1=46\)
- \(f_0=38\)
- \(f_2=34\)
- \(h=10\)
Substitution
\[ \text{Mode} = 20+ \left( \frac{46-38}{2(46)-38-34} \right)\times10 \]
\[ = 20+ \left( \frac{8}{92-72} \right)\times10 \]
\[ = 20+ \left( \frac8{20} \right)\times10 \]
\[ = 20+4 \]
\[ =24 \]
✓ Mode:
\[
\boxed{24}
\]
| Weight (kg) | 25–34 | 35–44 | 45–54 | 55–64 | 65–74 | 75–84 |
|---|---|---|---|---|---|---|
| No. of students | 4 | 8 | 10 | 14 | 8 | 6 |
Modal class
Highest frequency:
\[
14
\]
Thus modal class:
\[
55-64
\]
Formula
\[ \text{Mode} = l+ \left( \frac{f_1-f_0}{2f_1-f_0-f_2} \right)\times h \]
Where:
- \(l=55\)
- \(f_1=14\)
- \(f_0=10\)
- \(f_2=8\)
- \(h=10\)
Substitution
\[ \text{Mode} = 55+ \left( \frac{14-10}{2(14)-10-8} \right)\times10 \]
\[ = 55+ \left( \frac4{28-18} \right)\times10 \]
\[ = 55+ \left( \frac4{10} \right)\times10 \]
\[ = 55+4 \]
\[ =59 \]
✓ Mode:
\[
\boxed{59\text{ kg}}
\]
Statistics – Validated & Corrected Answers
Find lower limit.
Midpoint formula:
:contentReference[oaicite:0]{index=0}
Where:
- \(l\) = lower limit
- \(b\) = upper limit
Solving:
\[ 2m=l+b \]
\[ l=2m-b \]
✓ Answer:
\[
\boxed{(1)\ 2m-b}
\]
One number discarded, mean of remaining 6 numbers is 78.
Find discarded number.
Sum of 7 numbers
\[ 7\times81=567 \]
Sum of remaining 6 numbers
\[ 6\times78=468 \]
Discarded number:
\[ 567-468=99 \]
✓ Answer:
\[
\boxed{(3)\ 99}
\]
This is the definition of mode.
✓ Answer:
\[
\boxed{(3)\ \text{Mode}}
\]
Check option (2):
\[ 1,\ 3,\ 3,\ 3,\ 5 \]
Mean:
\[ \frac{1+3+3+3+5}{5} = \frac{15}{5} = 3 \]
Median:
\[
3
\]
Mode:
\[
3
\]
All equal.
✓ Answer:
\[
\boxed{(2)\ 1,3,3,3,5}
\]
Property:
\[ \sum (x-\bar{x})=0 \]
✓ Answer:
\[
\boxed{(1)\ 0}
\]
Mean of \(a,c,e\) is 24.
Find mean of \(b,d\).
Total of 5 numbers
\[ 5\times28=140 \]
Sum of \(a,c,e\)
\[ 3\times24=72 \]
Thus:
\[ b+d=140-72 \]
\[ =68 \]
Mean of \(b,d\):
\[ \frac{68}{2}=34 \]
✓ Answer:
\[
\boxed{(4)\ 34}
\]
Sum of five terms = 5×11 = 55. Sum = x+(x+2)+(x+4)+(x+6)+(x+8)=5x+20.
So 5x+20=55 ⇒ 5x=35 ⇒ x=7.
First three observations: 7, 9, 11. Their mean = (7+9+11)/3 = 27/3 = 9.
Sum of the five numbers = 5×13 = 65.
Sum of known numbers = 5+9+17+21 = 52.
Thus x = 65 − 52 = 13.
Squares:
\[ 1^2,2^2,3^2,\dots,11^2 \]
Sum formula:
:contentReference[oaicite:1]{index=1}
For \(n=11\):
\[ \sum x^2 = \frac{11(12)(23)}{6} \]
\[ =506 \]
Mean:
\[ \frac{506}{11} = 46 \]
✓ Answer:
\[
\boxed{(2)\ 46}
\]
If each number is multiplied by \(z\), new mean is
Property of mean:
- Multiplying each observation by \(z\) multiplies mean by \(z\).
New mean:
\[ z\bar{x} \]
✓ Answer:
\[
\boxed{(3)\ z\bar{x}}
\]
Answer Key
| Question | Answer |
|---|---|
| 1 | (1) |
| 2 | (3) |
| 3 | (3) |
| 4 | (2) |
| 5 | (1) |
| 6 | (4) |
| 7 | (1) |
| 8 | (2) |
| 9 | (2) |
| 10 | (3) |
Project Ideas
Steps:
- Collect speeds of 20 land animals
- Prepare frequency table
- Find:
- Mean
- Median
- Mode
- Compare and justify which measure best represents the data
Using class records:
(i) Mean age of class
- Group ages into intervals
- Use grouped mean formula
(ii) Mean height of class
- Group heights into intervals
- Calculate grouped mean
---# UNIT 9 : Probability
Validated & Corrected Answers
Probability Formula
:contentReference[oaicite:0]{index=0}
A week has:
\[
7
\]
days.
Only one favorable day:
\[
\text{Sunday}
\]
Thus:
\[ P(\text{Sunday}) = \frac17 \]
✓ Answer:
\[
\boxed{\frac17}
\]
A standard deck has:
\[
52
\]
cards.
Number of:
- Kings = 4
- Queens = 4
- Jacks = 4
Total favorable cards:
\[ 4+4+4=12 \]
Probability:
\[ P = \frac{12}{52} = \frac3{13} \]
✓ Answer:
\[
\boxed{\frac3{13}}
\]
Possible outcomes:
\[
1,2,3,4,5,6
\]
Even numbers:
\[
2,4,6
\]
Favorable outcomes:
\[
3
\]
Total outcomes:
\[
6
\]
\[ P = \frac36 = \frac12 \]
✓ Answer:
\[
\boxed{\frac12}
\]
Total balls:
\[
24
\]
- Red = 3
- Blue = 5
- Green:
\[ 24-3-5=16 \]
(i) Probability of Blue ball
\[ P(\text{Blue}) = \frac5{24} \]
✓ Answer:
\[
\boxed{\frac5{24}}
\]
(ii) Probability of Red ball
\[ P(\text{Red}) = \frac3{24} = \frac18 \]
✓ Answer:
\[
\boxed{\frac18}
\]
(iii) Probability of Green ball
\[ P(\text{Green}) = \frac{16}{24} = \frac23 \]
✓ Answer:
\[
\boxed{\frac23}
\]
Sample space:
\[ HH,\ HT,\ TH,\ TT \]
Total outcomes:
\[
4
\]
Favorable outcome:
\[
HH
\]
Thus:
\[ P = \frac14 \]
✓ Answer:
\[
\boxed{\frac14}
\]
Total outcomes:
\[ 6\times6=36 \]
(i) Sum equal to 1
Minimum possible sum:
\[
2
\]
So sum 1 is impossible.
\[ P=0 \]
✓ Answer:
\[
\boxed{0}
\]
(ii) Sum equal to 4
Favorable outcomes:
\[ (1,3),(2,2),(3,1) \]
Number of favorable outcomes:
\[
3
\]
Probability:
\[ \frac3{36} = \frac1{12} \]
✓ Answer:
\[
\boxed{\frac1{12}}
\]
(iii) Sum less than 13
Maximum possible sum:
\[
12
\]
All outcomes satisfy condition.
\[ P = \frac{36}{36} = 1 \]
✓ Answer:
\[
\boxed{1}
\]
Total LEDs tested:
\[
7000
\]
Defective LEDs:
\[
25
\]
Probability:
\[ P = \frac{25}{7000} \]
\[ = \frac1{280} \]
✓ Answer:
\[
\boxed{\frac1{280}}
\]
Successful goals:
\[ 40-32=8 \]
Probability of scoring goal:
\[ P = \frac8{40} = \frac15 \]
✓ Answer:
\[
\boxed{\frac15}
\]
From the spinner:
- Total sectors = 8
- Multiples of 3 = 2 sectors
Non-multiples of 3:
\[ 8-2=6 \]
Probability:
\[ \frac68 = \frac34 \]
✓ Answer:
\[
\boxed{\frac34}}
\]
Example Problem 1: What is the probability that the spinner lands on an even number?
Example Problem 2: What is the probability that the spinner lands on a prime number?
Total laptops manufactured:
\[
10000
\]
Defective laptops:
\[
25
\]
Good laptops:
\[ 10000-25=9975 \]
Probability of selecting a good laptop:
\[ P(\text{good}) = \frac{9975}{10000} \]
\[ =0.9975 \]
✓ Answer:
\[
\boxed{0.9975}
\]
Total youngsters surveyed:
\[
400
\]
Youngsters having voter ID:
\[
191
\]
Youngsters without voter ID:
\[ 400-191=209 \]
Probability:
\[ P(\text{No voter ID}) = \frac{209}{400} \]
✓ Answer:
\[
\boxed{\frac{209}{400}}
\]
Solution. The probabilities of an event and its complement add to 1, so
x/3 + x/5 = 1.
LCM of 3 and 5 is 15: (5x + 3x)/15 = 1 ⇒ 8x/15 = 1 ⇒ 8x = 15 ⇒ x = 15/8.
Answer: 15/8.
Find probability of losing.
Using:
\[ P(\text{lose})=1-P(\text{win}) \]
\[ =1-0.72 \]
\[ =0.28 \]
✓ Answer:
\[
\boxed{0.28}
\]
From the table:
- Families with both maids = 250
- Families with part-time maids = 860
- Families with no maids = 20
- Total families = 1500
(i) Probability of both types of maids
\[ P = \frac{250}{1500} \]
\[ = \frac16 \]
✓ Answer:
\[
\boxed{\frac16}
\]
(ii) Probability of part-time maids
\[ P = \frac{860}{1500} \]
\[ = \frac{43}{75} \]
✓ Answer:
\[
\boxed{\frac{43}{75}}
\]
(iii) Probability of no maids
\[ P = \frac{20}{1500} \]
\[ = \frac1{75} \]
✓ Answer:
\[
\boxed{\frac1{75}}
\]
Final Answers
| Question | Answer |
|---|---|
| 1 | \(0.9975\) |
| 2 | \(\frac{209}{400}\) |
| 3 | \(\frac{15}{8}\) |
| 4 | \(0.28\) |
| 5(i) | \(\frac16\) |
| 5(ii) | \(\frac{43}{75}\) |
| 5(iii) | \(\frac1{75}\) |
---Exercise 9.3
Multiple choice questions
1. A number between 0 and 1 that is used to measure uncertainty is called
(1) Random variable
(2) Trial
(3) Simple event
(4) Probability
[Answer: (4) Probability]
2. Probability lies between
(1) −1 and +1
(2) 0 and 1
(3) 0 and n
(4) 0 and ∞
[Answer: (2) 0 and 1]
3. The probability based on the concept of relative frequency theory is called
(1) Empirical probability
(2) Classical probability
(3) Both (1) and (2)
(4) Neither (1) nor (2)
[Answer: (1) Empirical probability]
4. The probability of an event cannot be
(1) Equal to zero
(2) Greater than zero
(3) Equal to one
(4) Less than zero
[Answer: (4) Less than zero]
5. The probability of all possible outcomes of a random experiment is always equal to
(1) One
(2) Zero
(3) Infinity
(4) Less than one
[Answer: (1) One]
6. If A is any event in S and its complement is A’ then, P(A′) is equal to
(1) 1
(2) 0
(3) 1−A
(4) 1−P(A)
[Answer: (4) 1−P(A)]
7. Which of the following cannot be taken as probability of an event?
(1) 0
(2) 0.5
(3) 1
(4) −1
[Answer: (4) −1]
8. A particular result of an experiment is called
(1) Trial
(2) Simple event
(3) Compound event
(4) Outcome
[Answer: (4) Outcome]
9. A collection of one or more outcomes of an experiment is called
(1) Event
(2) Outcome
(3) Sample point
(4) None of the above
[Answer: (1) Event]
10. The six faces of the dice are called equally likely if the dice is
(1) Small
(2) Fair
(3) Six-faced
(4) Round
Solution: Fair means all outcomes are equally likely.
[Answer: (2) Fair]
Exercise 9.3
1. (4) 2. (2) 3. (1) 4. (4) 5. (1) 6. (4) 7. (4) 8. (4) 9. (1) 10. (2)
Revise Statistics with confidence.
Use these expandable answers for quick revision, homework checking, and exam preparation.