Standard Deviation MCQ Quiz in தமிழ் - Objective Question with Answer for Standard Deviation - இலவச PDF ஐப் பதிவிறக்கவும்

The correct answer is 50.00.

Key Points 

Firstly we will have to standardise the given values so that a standard normal table can be used for finding the probability.

In this question, we have Mean(X) = 120.

                                          Standard Deviation () = 35.

We have to use this formula: Z = X-µ/ð

  Z₁ = 85 - 120/35 = -2.714

  Z₂ = 155 -120/35 = 1

Now, we will have to use the standard normal table to find area between -2.714 and 1.

From the above table, we can find out the exact probability.

Probability= 0.00336-0.84134=0.83=83%

Important Points 

Standard deviation:

A standard deviation is a statistical tool that measures the dispersion relative to the mean. It is calculated by finding the square root of variance. It is denoted by ð. Standard deviation in finance is used for measuring the riskiness of the asset.

The formula for standard deviation is:

ð = \(\sqrt{\dfrac{\sum\limits_{}^{}{{\lvert x-\mu\rvert^2}}}{N}}\)

Where, ð= Standard deviation.

  N= Population

  μ = Population Mean

           x = Observations

• It helps in identifying the place where the data is located and dispersed in a frequency distribution.
• It helps in evaluating how far the observation is distributed away from the mean.
• It includes all types of data in observation.

Hence, the correct answer is 50.00

The correct answer is the product of the standard deviation of X and Y (SxSy).

Key Points Standard Deviation:

The equation for the Pearson correlation coefficient "r" involves dividing the covariance between variables X and Y (COVxy) by the product of their standard deviations (SxSy). The formula for calculating "r" is:

r = COVxy / (Sx * Sy)

Where:

• COVxy represents the covariance between variables X and Y
•  represents the standard deviation of variable X
• Sy represents the standard deviation of variable Y

The Pearson correlation coefficient "r" is a measure of the strength and direction of the linear relationship between two variables.

The value of "r" ranges from -1 to +1, where -1 indicates a perfect negative correlation, +1 indicates a perfect positive correlation, and 0 indicates no correlation.

By dividing the covariance between variables X and Y by the product of their standard deviations, the Pearson correlation coefficient "r" is able to normalize the relationship between the variables, making it easier to compare and interpret the strength and direction of the relationship, regardless of the units or scales of the variables.

Hence, the correct answer is COVxy; the product of the standard deviation of X and Y (SxSy).

Explanation:

Suppose the values in the first group are x1i, i = 1,2,……,n1 and those in the second group are x2i, i = 1,2,……..,n2.

Now since x̅1 and x̅2 are the means so, x̅1=\(\frac{1}{{{n_1}}}\mathop \sum \limits_{i = 1}^{{n_1}} {x_{1i}}\) and  x̅2 =\(\frac{1}{{{n_2}}}\mathop \sum \limits_{i = 1}^{{n_2}} {x_{2i}}\)  

If the mean of the combined series is x̅, then x̅ = \(\frac{{{n_1}\overline {{x_1}} + {n_2}\overline {{x_2}} }}{{{n_1} + {n_2}}}\)

Again σ1 and σ2 are the standard deviations. So, variance \({\rm{\sigma }}_1^2 = \frac{1}{{{n_1}}}\mathop \sum \limits_{i = 1}^{{n_1}} {\left( {{x_{1i}} - {{{\rm{\bar x}}}_1}{\rm{\;}}} \right)^2}\;\) and \({\rm{\sigma }}_2^2 = \frac{1}{{{n_2}}}\mathop \sum \limits_{i = 1}^{{n_2}} {\left( {{x_{2i}} - {{{\rm{\bar x}}}_2}{\rm{\;}}} \right)^2}.\)

Now, the sum of squares of the deviation of the values of the two groups from x̅ is:

\(\Rightarrow\mathop \sum \limits_{i = 1}^{{n_1}} {\left( {{x_{2i}} - \bar x} \right)^2} + \mathop \sum \limits_{i = 1}^{{n_2}} {\left( {{x_{2i}} - \overline {x{\rm{\;}}} } \right)^2}\)

Now,

\(\mathop \sum \limits_{i = 1}^{{n_1}} {\left( {{x_{1i}} - \bar x} \right)^2}\;=\mathop \sum \limits_{i = 1}^{{n_1}} \;{\left\{ {\left( {{x_{1i}} - {\rm{\;}}\overline {{x_1}} } \right) + \left( {\overline {{x_1}} - \bar x{\rm{\;}}} \right)} \right\}^2}\)

\(=\mathop \sum \limits_{i = 1}^{{n_1}} {\left( {{x_{1i}} - {{{\rm{\bar x}}}_1}{\rm{\;}}} \right)^2}\;+2\left( {\overline {{x_i}} - \bar x} \right)\mathop \sum \limits_{i = 1}^{{n_1}} \left( {{x_{1i}} - \overline {{x_1}} } \right) + {n_1}{\left( {\overline {{x_1}} - \bar x} \right)^2}\)

\( = {n_1}\sigma _1^2 + \;{n_1}{\left( {\overline {{x_1}} - \bar x} \right)^2},\;\;\;\mathop \sum \limits_{i = 1}^{{n_1}} \left( {{x_{1i}} - \overline {{x_1}} } \right) = 0\)

Similarly, \(\mathop \sum \limits_{i = 1}^{{n_2}} {\left( {{x_{2i}} - \overline {x{\rm{\;}}} } \right)^2} = {n_2}\sigma _2^2 + {n_2}{\left( {\overline {{x_2}} - \bar x} \right)^2}\)

If the variation of the combined series is σ2,

Then, \({\sigma ^2} = \frac{1}{{{n_1} + {n_2}}}\left\{ {\mathop \sum \limits_{i = 1}^{{n_1}} {{\left( {{x_{2i}} - \bar x} \right)}^2}\; + {\rm{\;}}\mathop \sum \limits_{i = 1}^{{n_2}} {{\left( {{x_{2i}} - \overline {x{\rm{\;}}} } \right)}^2}} \right\}\)

\( = \frac{{{n_1}\sigma _1^2 + {n_2}\sigma _2^2}}{{{n_1} + {n_2}}}+\frac{{{n_1}{{\left( {\overline {{x_1}} - \bar x} \right)}^2} + {n_2}{{\left( {\overline {{x_2}} - \bar x} \right)}^2}}}{{{n_1} + {n_2}}}\)

\(=\frac{{{n_1}\sigma _1^2 + {n_2}\sigma _2^2}}{{{n_1} + {n_2}}}+\frac{{{n_1}d_1^2 + {n_2}d_2^2}}{{{n_1} + {n_2}}}\)

\(=\frac{1}{{{n_1} + {n_2}}}\left( {{n_1}\left( {\sigma _1^2 + d_1^2} \right) + {n_2}\left( {\sigma _2^2 + d_2^2} \right)} \right)\)

Standard deviation doesn't change when

- graph is shifted right or left

- graph is moved up or moved down

- graph is rotated around any point.

Standard deviation may change when the graph is sketched or squeezed.

If you observe carefully, both the distributions given in the question, the spread of the graph has been rotated from 30. Hence this would not change the standard deviation.