Jamia Millia Islamia MCA Statistics Measures Of Central Tendency Previous Year Questions (PYQs) – Page 1 of 2
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For individual observations, reciprocal of arithmetic mean is called
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Solution
Reciprocal of arithmetic mean = harmonic mean.
the standard deviation of some temperature data in °C is 5.
If the data were converted into °F, the variance would be:
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Solution
Conversion formula: $F = \dfrac{9}{5}C + 32$
So, new standard deviation = $\dfrac{9}{5} \times 5 = 9$
Variance = $(\text{SD})^2 = 9^2 = 81$
$\boxed{\text{Answer: (A) 81}}$
The mean of 100 observations is 50 and their standard deviation is 5.
The sum of squares of all observations is —
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Solution
Let $\bar{x} = 50$, $\sigma = 5$, and $n = 100.$
We know,
$\sigma^2 = \dfrac{\sum x^2}{n} - \bar{x}^2$
$\Rightarrow \sum x^2 = n(\sigma^2 + \bar{x}^2)$
$= 100(5^2 + 50^2) = 100(25 + 2500) = 100 \times 2525 = 2,52,500.$
The mean and standard deviation of marks obtained by 50 students of a cl ass i n
three subjects physics, mathematics and chemistry are as follows:
Which of the subjects show the highest and lowest variabilities respectively?
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Solution
Compare variability via coefficient of variation (CV) = SD / Mean.
Math: $12/42 \approx 0.286$
Physics: $15/32 \approx 0.469$
Chemistry: $20/40.9 \approx 0.489$
Highest CV → Chemistry; Lowest CV → Mathematics.
$\boxed{\text{Answer: (B) Chemistry, Mathematics}}$
What will be the mean and variance for the first $n$ natural numbers?
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Solution
For $1,2,\dots,n$:
Mean $= \dfrac{1+\cdots+n}{n} = \dfrac{n(n+1)/2}{n} = \dfrac{n+1}{2}$.
Variance $= \dfrac{1}{n}\sum_{k=1}^{n}\left(k-\dfrac{n+1}{2}\right)^{2}
= \dfrac{n^{2}-1}{12}$.
$\boxed{\dfrac{n+1}{2},\ \dfrac{n^{2}-1}{12}}$
The median of a set of $9$ distinctive observations is $20.5$. If each of the largest $4$ observations of the set is increased by $2$, then the median of the new set
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Solution
For $9$ observations, median is the $5^{\text{th}}$ term. Increasing the top $4$ (positions $6$–$9$) does not change the $5^{\text{th}}$ value.
Marks obtained by $4$ students are: $25,35,45,55$. The average deviation from the mean is
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Solution
$= \dfrac{25+35+45+55}{4}=40$.
Absolute deviations: $|25-40|,|35-40|,|45-40|,|55-40|=15,5,5,15$.
Average deviation $=\dfrac{15+5+5+15}{4}=10$.
The numbers $3, 5, 7, 4$ have frequencies $x, x+4, x-3, x+8$.
If their arithmetic mean is $4$, then the value of $x$ is:
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Solution
$\text{Mean} = \dfrac{3x + 5(x+4) + 7(x-3) + 4(x+8)}{x + (x+4) + (x-3) + (x+8)} = 4$
$\Rightarrow \dfrac{19x + 33}{4x + 9} = 4$
$\Rightarrow 19x + 33 = 16x + 36$
$\Rightarrow 3x = 3$
$\Rightarrow x = 1$
If A.M. of two numbers is $15/2$ and their G.M. is $6$, then find the two numbers.
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Solution
Let the numbers be $a$ and $b$.
$\dfrac{a + b}{2} = \dfrac{15}{2} \Rightarrow a + b = 15$
and $\sqrt{ab} = 6 \Rightarrow ab = 36$
Now, $a$ and $b$ are roots of $x^2 - 15x + 36 = 0$
$\Rightarrow x = 12, 3$
Hence, the numbers are 12 and 3.
The function $f:[0,3] \to [1,29]$ defined by
$f(x) = 2x^3 - 15x^2 + 36x + 1$ is:
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Solution
$f'(x) = 6x^2 - 30x + 36 = 6(x^2 - 5x + 6) = 6(x-2)(x-3).$
Sign chart:
• $f'$ > 0 for $x<2$; $f'$ < 0 for $2
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