JEE Mathematics Statistics MCQs Set 01

Answer: (c) \( \frac53 \)

Explanation:
1. Let the groups have n₁ and n₂ persons: \( \frac{400n_1+480n_2}{n_1+n_2}=430 \).
2. \( 400n_1+480n_2=430n_1+430n_2\Rightarrow50n_2=30n_1 \), so \( \frac{n_1}{n_2}=\frac53 \).

Teacher's Note:
1. Equate the combined total income to 430 times the total number of persons.
2. Rearrange to get the ratio.

Answer: (b) \( \bar x+\lambda \)

Explanation:
1. The new sum is \( \sum x_i+n\lambda \).
2. Dividing by n gives \( \bar x+\lambda \).

Teacher's Note:
1. Adding a constant to every observation adds it to the mean.
2. The sum increases by nλ.

Answer: (b) 79.48

Explanation:
1. Original sum \( =25\times78.4=1960 \).
2. Correct sum \( =1960-69+96=1987 \).
3. Correct mean \( =\frac{1987}{25}=79.48 \).

Teacher's Note:
1. Correct the total, not the mean directly.
2. Replace 69 by 96 in the sum.

Answer: (c) Rs. 891.2

Explanation:
1. Total workers \( =7+14+19+25+20+10+5=100 \).
2. \( \sum fx=5600+11480+16340+22500+18400+9800+5000=89120 \).
3. Mean \( =\frac{89120}{100}=891.2 \).

Teacher's Note:
1. Use the weighted mean Σfx/Σf.
2. Check that the frequencies add up to 100.

Answer: (b) 3.16

Explanation:
1. \( H=\frac{3}{\frac12+\frac14+\frac15}=\frac{3}{0.95}\approx3.16 \).

Teacher's Note:
1. HM = n divided by the sum of reciprocals.
2. 1/2 + 1/4 + 1/5 = 0.95.

Answer: (b) 27

Explanation:
1. Arrange in order: 0, 5, 11, 19, 21, 27, 30, 36, 42, 50, 52.
2. With 11 values, the median is the 6th value: 27.

Teacher's Note:
1. Sort the data first.
2. For n = 11 the median is the ((n + 1)/2)-th value.

Answer: (b) 5

Explanation:
1. Total frequency 120; cumulative frequencies: 8, 18, 29, 45, 65, 90, 105, 114, 120.
2. The median lies between the 60th and 61st observations, both of which lie in the cumulative class 45–65, i.e. x = 5.

Teacher's Note:
1. Find the cumulative frequencies.
2. Locate the observations numbered N/2 and N/2 + 1.

Answer: (d) 19.5

Explanation:
1. N = 49, so N/2 = 24.5. Cumulative frequencies: 5, 11, 26, ... so the median class is 15–20.
2. \( \text{Median}=l+\frac{N/2-F}{f}\times h=15+\frac{24.5-11}{15}\times5=15+4.5=19.5 \).

Teacher's Note:
1. Find the class in which the cumulative frequency first exceeds N/2.
2. Apply the median formula for grouped data.

Answer: (c) 5

Explanation:
1. Counts: 2 appears 3 times, 3 appears 3 times, 5 appears 4 times, 6 appears 2 times.
2. The most frequent value is 5.

Teacher's Note:
1. Tabulate the frequency of each value.
2. The mode is the value with the highest frequency.

Answer: (b) 64

Explanation:
1. Empirical relation: \( \text{Mode}=3\,\text{Median}-2\,\text{Mean} \).
2. \( 60=3M-132\Rightarrow M=64 \).

Teacher's Note:
1. Use Mode = 3 Median − 2 Mean.
2. Solve for the median.

Answer: (b) 7.09

Explanation:
1. N = 44; \( \sum fx=24+90+204+207+135=660 \), mean 15.
2. Deviations \( |x-15| \): 12, 6, 2, 8, 12; \( \sum f|x-15|=96+60+24+72+60=312 \).
3. \( \text{MD}=\frac{312}{44}\approx7.09 \).

Teacher's Note:
1. Compute the mean first.
2. MD = Σf|x − mean|/N.

Answer: (b) 21.2

Explanation:
1. Mean = 70/5 = 14. Squared deviations: 36, 4, 1, 1, 64; sum 106.
2. Variance \( =\frac{106}5=21.2 \).

Teacher's Note:
1. Variance = mean of squared deviations.
2. Compute the mean first.

Answer: (c) 1.32

Explanation:
1. N = 217. Mean \( =\frac{\sum fx}{N}=\frac{1538.5}{217}\approx7.09 \).
2. \( \sigma^2=\frac{\sum fx^2}{N}-\bar x^2\approx1.32 \).

Teacher's Note:
1. Use σ² = Σfx²/N − (mean)².
2. Keep several decimals while computing the mean.

Answer: (d) 18

Explanation:
1. \( np=6 \) and \( npq=4 \), so \( q=\frac23,\ p=\frac13 \).
2. \( n=\frac{6}{1/3}=18 \).

Teacher's Note:
1. Divide variance by mean to get q.
2. Then n = mean/p.