Statistics and Random Variables: questions with solutions
3 free questions from this chapter (the full set has 30). Marking: Multi-correct: +4 all correct, +1 per correct option if none wrong, −2 if any wrong · Numerical: +4 / 0 · Matching: +3 / −1.
Question 1Level 2 · ExamSingle correct
The frequency distribution below has total frequency 60:
classes 0–10, 10–20, 20–30, 30–40, 40–50 with frequencies 5,f1,20,f2,5 respectively, where f1,f2 are non-negative integers. The median of the distribution is 28. Each class a–b contains a but not b; use class marks for the mean and the standard interpolation formulas: median =l+fN/2−Ch and mode =l+2f1−f0−f2f1−f0h. The mode of the distribution is ______. Exactly one option is correct.
A31
B30
C17520
D27
Show answer and solution
Answer: C
5+f1+20+f2+5=60 gives f1+f2=30.
Median 28 lies in 20–30, so C=5+f1, f=20, h=10: 28=20+2030−(5+f1)⋅10, i.e. 225−f1=8, so f1=9 and f2=21. (Check: cumulative frequency before 20 is 14<30≤34, so 20–30 is indeed the median class.)
Now the largest frequency is f2=21, so the modal class is 30–40, not 20–30: mode =30+2(21)−20−521−20⋅10=30+1710=17520.
Trap: using 20–30 as modal class gives 20+1011⋅10=31, which is not even inside that class; the empirical relation 3(28)−2(27)=30 (mean is 27) is only an approximation.
Difficulty: Hard
Question 2Level 2 · ExamSingle correct
For 20 observations, the mean and variance were computed as 10 and 9. It was later found that one observation recorded as 15 was actually 25. After this correction, every observation x is replaced by 4−2x. Variance is σ2=N1∑fi(xi−xˉ)2 with divisor N=∑fi (total frequency), and the standard deviation (SD) is σ≥0. The standard deviation of the new data is ______. Exactly one option is correct.
A53
B229
C219
D77
Show answer and solution
Answer: A
Wrong data: ∑x=200, ∑x2=20(9+100)=2180.
Corrected: ∑x=210, ∑x2=2180−225+625=2580, so xˉ=10.5 and σ2=202580−10.52=129−110.25=18.75.
For y=4−2x: σy=∣−2∣σx=218.75=2⋅253=53.
Trap: keeping the old mean 10 gives σ2=29 and the option 229.
Difficulty: Hard
Question 3Level 2 · ExamSingle correct
For n observations x1,…,xn it is known that ∑i=1n(xi−1)=10, ∑i=1n(xi−1)2=40 and the variance is 3. Variance is σ2=N1∑fi(xi−xˉ)2 with divisor N=∑fi (total frequency), and the standard deviation (SD) is σ≥0. The value of ∑i=1nxi2 is ______. Exactly one option is correct.
A80
B60
C50
D70
Show answer and solution
Answer: D
Variance is unchanged by the shift ui=xi−1: 3=n40−(n10)2, so 3n2−40n+100=0, (n−10)(3n−10)=0. Since n is a positive integer, n=10.
Then ∑xi=10+n=20 and ∑xi2=∑(xi−1)2+2∑xi−n=40+40−10=70.
Trap: forgetting the −n term gives 80; using ∑(xi−1)=10 in place of ∑xi gives 50.
Difficulty: Hard
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