Qus : 7 Phrases PYQ 4 If ${{a}}_1,{{a}}_2,\ldots,{{a}}_n$ are any real numbers and $n$ is
any positive integer, then
1 $$n\sum ^n_{i=1}{{{a}}_i}^2{\lt}{\Bigg{(}\sum ^n_{i=1}{{{a}}_i}\Bigg{)}}^2$$ 2 $$n\sum ^n_{i=1}{{{a}}_i}^2{\geq}{\Bigg{(}\sum ^n_{i=1}{{{a}}_i}\Bigg{)}}^2$$ 3 $$\sum ^n_{i=1}{{{a}}_i}^2{\geq}{\Bigg{(}\sum ^n_{i=1}{{{a}}_i}\Bigg{)}}^2$$ 4 None of the above Go to Discussion Phrases Previous Year PYQ Phrases NIMCET 2022 PYQ
Solution Qus : 12 Phrases PYQ 2 If a variable takes values 0, 1, 2,…, 50 with frequencies $1,\, {{50}}_{{{C}}_1},{{50}}_{{{C}}_2},\ldots..,{{50}}_{{{C}}_{50}}$, then the AM is
1 50 2 25 3 $\frac{{2}^{50}}{50}$ 4 51 Go to Discussion Phrases Previous Year PYQ Phrases NIMCET 2021 PYQ
Solution Qus : 14 Phrases PYQ 4 A, B, C are three sets of values of x:
A: 2,3,7,1,3,2,3
B: 7,5,9,12,5,3,8
C: 4,4,11,7,2,3,4
Select the correct statement among the following:
1 Mean of A is equal to Mode of C 2 Mean of C is equal to Median of B 3 Median of B is equal to Mode of A 4 Mean, Median and Mode of A are same Go to Discussion Phrases Previous Year PYQ Phrases NIMCET 2020 PYQ Solution A: 2, 3, 7, 1, 3, 2, 3
Increasing Order : A: 1, 2, 2, 3, 3, 3, 7
Mode = 3 (occurs maximum number of times)
Median = 3 (the middle term)
Mean =$\frac{(1+2+2+3+3+3+7)}{7}$
$=\frac{21}{7} = 3$
Hence. Mean=Median=Mode
Qus : 15 Phrases PYQ 1 Standard deviation for the following distribution is
Size of item 6 7 8 9 10 11 12 Frequency 3 6 9 13 8 5 4
1 1.607 2 9.0 3 5.0 4 1.88 Go to Discussion Phrases Previous Year PYQ Phrases NIMCET 2020 PYQ Solution Total number of items in the distribution = Σ fi = 3 + 6 + 9 + 13 + 8 + 5 + 4 = 48.
The Mean (x̅) of the given set = \(\rm \dfrac{\sum f_i x_i}{\sum f_i}\) .
⇒ x̅ = \(\rm \dfrac{6\times3+7\times6+8\times9+9\times13+10\times8+11\times5+12\times4}{48}=\dfrac{432}{48}\) = 9.
Let's calculate the variance using the formula: \(\rm \sigma^2 =\dfrac{\sum x_i^2}{n}-\bar x^2\) .
\(\rm \dfrac{\sum {x_i}^2}{n}=\dfrac{6^2\times3+7^2\times6+8^2\times9+9^2\times13+10^2\times8+11^2\times5+12^2\times4}{48}=\dfrac{4012}{48}\) = 83.58.
∴ σ2 = 83.58 - 92 = 83.58 - 81 = 2.58.
And, Standard Deviation (σ) = \(\rm \sqrt{\sigma^2}=\sqrt{Variance}=\sqrt{2.58}\) ≈ 1.607 .
Qus : 17 Phrases PYQ 3 For a group of 100 candidates, the mean and standard deviation of scores were found to be 40 and 15
respectively. Later on, it was found that the scores 25 and 35 were misread as 52 and 53 respectively. Then the
corrected mean and standard deviation corresponding to the corrected figures are
1 39.9, 14.97 2 39.5, 14 3 39.55, 14.97 4 40.19, 15.1 Go to Discussion Phrases Previous Year PYQ Phrases NIMCET 2023 PYQ
Solution [{"qus_id":"3754","year":"2018"},{"qus_id":"3895","year":"2019"},{"qus_id":"3936","year":"2019"},{"qus_id":"3928","year":"2019"},{"qus_id":"3927","year":"2019"},{"qus_id":"9439","year":"2020"},{"qus_id":"9440","year":"2020"},{"qus_id":"10676","year":"2021"},{"qus_id":"10694","year":"2021"},{"qus_id":"10711","year":"2021"},{"qus_id":"11113","year":"2022"},{"qus_id":"11126","year":"2022"},{"qus_id":"11132","year":"2022"},{"qus_id":"11139","year":"2022"},{"qus_id":"11503","year":"2023"},{"qus_id":"11506","year":"2023"},{"qus_id":"11507","year":"2023"},{"qus_id":"11652","year":"2024"}]