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Jamia Millia Islamia Previous Year Questions (PYQs)

Jamia Millia Islamia Statistics Measures Of Central Tendency PYQ

Jamia Millia Islamia PYQ
For individual observations, reciprocal of arithmetic mean is called





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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2018 PYQ

Solution

Reciprocal of arithmetic mean = harmonic mean.
Jamia Millia Islamia PYQ
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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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2021 PYQ

Solution

Conversion formula: $F = \dfrac{9}{5}C + 32$ So, new standard deviation = $\dfrac{9}{5} \times 5 = 9$ Variance = $(\text{SD})^2 = 9^2 = 81$ $\boxed{\text{Answer: (A) 81}}$
Jamia Millia Islamia PYQ
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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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2023 PYQ

Solution

Let $\bar{x} = 50$, $\sigma = 5$, and $n = 100.$ We know, $\sigma^2 = \dfrac{\sum x^2}{n} - \bar{x}^2$ $\Rightarrow \sum x^2 = n(\sigma^2 + \bar{x}^2)$ $= 100(5^2 + 50^2) = 100(25 + 2500) = 100 \times 2525 = 2,52,500.$
Jamia Millia Islamia PYQ
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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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2020 PYQ

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. $\boxed{\text{Answer: (B) Chemistry, Mathematics}}$
Jamia Millia Islamia PYQ
What will be the mean and variance for the first $n$ natural numbers?





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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2020 PYQ

Solution

For $1,2,\dots,n$: Mean $= \dfrac{1+\cdots+n}{n} = \dfrac{n(n+1)/2}{n} = \dfrac{n+1}{2}$. Variance $= \dfrac{1}{n}\sum_{k=1}^{n}\left(k-\dfrac{n+1}{2}\right)^{2} = \dfrac{n^{2}-1}{12}$. $\boxed{\dfrac{n+1}{2},\ \dfrac{n^{2}-1}{12}}$
Jamia Millia Islamia PYQ
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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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2022 PYQ

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.
Jamia Millia Islamia PYQ
Marks obtained by $4$ students are: $25,35,45,55$. The average deviation from the mean is





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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2022 PYQ

Solution

$= \dfrac{25+35+45+55}{4}=40$. Absolute deviations: $|25-40|,|35-40|,|45-40|,|55-40|=15,5,5,15$. Average deviation $=\dfrac{15+5+5+15}{4}=10$.
Jamia Millia Islamia PYQ
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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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2022 PYQ

Solution

$\text{Mean} = \dfrac{3x + 5(x+4) + 7(x-3) + 4(x+8)}{x + (x+4) + (x-3) + (x+8)} = 4$ $\Rightarrow \dfrac{19x + 33}{4x + 9} = 4$ $\Rightarrow 19x + 33 = 16x + 36$ $\Rightarrow 3x = 3$ $\Rightarrow x = 1$
Jamia Millia Islamia PYQ
If A.M. of two numbers is $15/2$ and their G.M. is $6$, then find the two numbers.





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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2024 PYQ

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.
Jamia Millia Islamia PYQ
The function $f:[0,3] \to [1,29]$ defined by $f(x) = 2x^3 - 15x^2 + 36x + 1$ is:





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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2019 PYQ

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
Jamia Millia Islamia PYQ
. Find the variance of the observation values taken in the lab: $4.2,\ 4.3,\ 4.0,\ 4.1$.





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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2024 PYQ

Solution

Mean $,\bar x=\dfrac{4.2+4.3+4.0+4.1}{4}=4.15$. Deviations: $0.05,,0.15,,-0.15,,-0.05$. Sum of squares $=0.0025+0.0225+0.0225+0.0025=0.05$. Population variance $,s^2=\dfrac{0.05}{4}=0.0125$. Sample variance $,s^2=\dfrac{0.05}{3}\approx0.0167$.
Jamia Millia Islamia PYQ
If the coefficient of variation is $100$ and the mean of the data is $25$, then find the standard deviation.





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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2024 PYQ

Solution

Coefficient of variation $=$ $\dfrac{\sigma}{\bar{x}}\times100$ $\Rightarrow 100 = \dfrac{\sigma}{25} \times 100$ $\Rightarrow \sigma = 25$
Jamia Millia Islamia PYQ
If mean is 11 and median is 13, then value of mode is





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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2018 PYQ

Solution

(Using the formula: Mode = 3 × Median − 2 × Mean = 3×13 − 2×11 = 39 − 22 = 17)

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