AMU MCA Statistics Measures of Central Tendency Previous Year Question Paper and Solution with PDF

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AMU MCA Statistics Measures Of Central Tendency PYQ

Qus : 1 AMU MCA PYQ

In the analysis of data of a randomized block design with 5 blocks and 4 treatments, the error degrees of freedom is:

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Solution

For Randomized Block Design:

Error d.f. = $(t-1)(b-1)$

Here $t=4$, $b=5$

Error d.f. = $(4-1)(5-1)=3 \times 4 = 12$

Qus : 2 AMU MCA PYQ

The standard deviation of sample means is called

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Solution

Standard deviation of sample mean distribution is called Standard Error.

Qus : 3 AMU MCA PYQ

Sum of the deviation about mean is

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Solution

By property of arithmetic mean:

$\sum (x_i-\bar{x})=0$

Qus : 4 AMU MCA PYQ

The standard deviation of a set of 50 observations is 8. If each observation is multiplied by 2, the new value of standard deviation will be

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Solution

If each observation is multiplied by a constant $k$, then new S.D. = $|k| \times$ old S.D.

Here $k=2$

New S.D. = $2 \times 8 = 16$

Qus : 5 AMU MCA PYQ

The standard deviation of first $n$ natural numbers is

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Solution

For numbers $1,2,\dots,n$:

Mean $=\frac{n+1}{2}$

Variance $=\frac{n^2-1}{12}$

S.D. $=\sqrt{\frac{n^2-1}{12}}$

Qus : 6 AMU MCA PYQ

Response error is a type of:

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Solution

Qus : 7 AMU MCA PYQ

If a random sample of size 20 from a normal population with variance 225 has mean $\bar{x}=64.3$, then the 95% confidence interval for the population mean $\mu$ is (Given that $Z_{0.025}=1.96$).

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Solution

Qus : 8 AMU MCA PYQ

Uniform distribution is a particular case of Beta distribution with parameters

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Solution

Qus : 9 AMU MCA PYQ

Suppose that there are $K$ treatments and $n$ blocks in a randomized block design. In analysis of variance for this design, the error degrees of freedom is:

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Solution

Qus : 10 AMU MCA PYQ

In $4 \times 4$ Latin square design, the error degrees of freedom is:

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Solution

Qus : 11 AMU MCA PYQ

In an analysis of variance, one estimate of variance $(\sigma^2)$ is based upon the difference between the treatment means and the

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Solution

Qus : 12 AMU MCA PYQ

Thirty men and thirty women with knee pain were subjects in an experiment to determine the effectiveness of a new pain medication. Fifteen of the 30 men and 15 of the 30 women were chosen randomly to receive the new drug. The remaining 15 men and 15 women received a placebo. The decrease in pain was measured for each subject. Then the design of experiment is

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Solution

Qus : 13 AMU MCA PYQ

The sampling scheme in which only first unit is selected randomly is known as:

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Solution

In systematic sampling, the first unit is selected randomly and the rest are selected at fixed intervals.

Qus : 14 AMU MCA PYQ

The empirical relationship between mean, median and mode is

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Solution

$Mode = 3Median − 2Mean$

Rearranging:

$Mean − Mode = 3(Mean − Median)$

Qus : 15 AMU MCA PYQ

A person covers 9 km at a speed of 3 km/hr, 25 km at a speed of 5 km/hr and 30 km at a speed of 10 km/hr. The average speed of the entire journey is:

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AMU MCA Previous Year PYQ AMU MCA AMU MCA 2024 PYQ

Solution

Total distance $= 9 + 25 + 30 = 64$ km Total time $= \frac{9}{3} + \frac{25}{5} + \frac{30}{10}$ $= 3 + 5 + 3 = 11$ hr Average speed $= \frac{64}{11} = 5\ \frac{9}{11}$ km/hr

Qus : 16 AMU MCA PYQ

A man purchased 5 toys at the rate of Rs. 200 each, 6 toys at the rate of Rs. 250 each and 9 toys at the rate of Rs. 300 each. The average cost of one toy is:

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Solution

$= 5 \times 200 + 6 \times 250 + 9 \times 300$ $= 1000 + 1500 + 2700 = 5200$ Total toys $= 5 + 6 + 9 = 20$ Average cost $= \frac{5200}{20} = 260$

Qus : 17 AMU MCA PYQ

In case of two-way classification with $r$ rows and $c$ columns, the degree of freedom for error is:

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Solution

In two-way ANOVA:

Qus : 18 AMU MCA PYQ

For a normal distribution, the coefficient of Kurtosis $\beta_2$ and $\gamma_2$ are:

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Solution

$ \beta_2=3 $

$ \gamma_2=\beta_2-3=0 $

Qus : 19 AMU MCA PYQ

First four moments about 5 are: $2,20,40,50$ Find S.D.

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Solution

Mean about 5: $\mu_1'=2$ Actual mean: $=5+2=7$ Variance: $\mu_2=20-2^2=16$ S.D.: $=\sqrt{16}=4$

Qus : 20 AMU MCA PYQ

Bowley’s coefficient of skewness is based on

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Solution

Bowley’s coefficient uses quartiles: $Sk = \dfrac{Q_3 + Q_1 - 2Q_2}{Q_3 - Q_1}$

Qus : 21 AMU MCA PYQ

If a constant value 5 is subtracted from each observation of a set, the variance is

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Solution

Variance is independent of change of origin.

Qus : 22 AMU MCA PYQ

Census survey is free from

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Solution

Census includes entire population → no sampling involved.

Qus : 23 AMU MCA PYQ

If experimental material is homogeneous, then suitable design is

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Solution

For homogeneous units, Completely Randomized Design (CRD) is best.

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