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Greater than or equal to

Explanation: 

We know standard deviation is difference between square of deviation of data about mean and square of mean.

And mean deviation is sum of all deviations of a set of data about the data's mean.

Hence, the standard deviation is greater than or equal to the mean deviation taken from the arithmetic mean.

Minimum

Explanation: 

The sum of the squares of the deviations of the values of the variable is minimum or least when taken about their arithmetic mean.

Correct option is (B) 3 2/3

Let sum of scored runs on first 5 balls in the last over is x.

After hitting 6 on last ball of the over, the mean score becomes 4 runs per ball.

\(\therefore\) \(\frac{x+6}6=4\)

\(\Rightarrow\) \(\frac{x}6+1=4\)

\(\Rightarrow\) \(\frac{x}6=4-1=3\)

\(\Rightarrow\) x = 6 \(\times\) 3 = 18

Hence, the sum of scored runs on first 5 balls in the last over is 18.

Given that Sachin scored 4 runs on last ball.

\(\therefore\) Mean \(=\frac{18+4}6\)

\(=\frac{22}6\) \(=3\frac46\)

\(=3\frac23\)

Hence, the mean score per ball in that over is \(3\frac23.\)

Correct option is  B) 3 2/3

Correct option is (C) 16

\(\overline X=\frac{\sum x_i}n\)

\(\Rightarrow\) \(n=\frac{\sum x_i}{\overline X}\)

\(=\frac{192}{12}\) = 16

Correct option is  C) 16

Grouped frequency table

Correct option is (B) 9

\(\because\) 9 occurs most often.

\(\therefore\) Mode of the given observations is 9.

Correct option is  B) 9

Correct option is (A) Range

Mean, median and mode is the measure of central tendency. But range is not a measure of central tendency.

Correct option is  A) Range

Correct option is (B) 14.5

Increased order of given observations is 10, 12, 14, 15, 16, 20.

Total number of observations is n = 6.

\(\therefore\) Median \(=\frac{\frac n2th+(\frac n2+1)th\text{ observations}}2\)

\(=\frac{(3^{rd}+4^{th})\text{ observations}}2\)    \((\because\) n = 6)

\(=\frac{14+15}2\)

= 14.5

Hence, median value of the given series is 14.5.

Correct option is  B) 14.5

No, since the number of trials in which the event can happen cannot be greater than the total number of trials.

No, since the number of trials in which the event can happen cannot be negative and the total number of trials is always positive.

Correct option is: C) a

Mean = \(\frac {sum \, of \, observations}{Total \, number \, of\, observations}\) 

= \(\frac {(a-2d)+(a-d)+(a+d)+(a+2d)}{4} = \frac {4a}4 =a\)

Correct option is: C) a

Correct option is: C) boundaries, cumulative frequency

While drawing ogive curve we take class boundaries on X-axis and cumulative frequency on Y-axis.

Correct option is: C) boundaries, cumulative frequency

Correct option is: A) Mean

To find the average of marks obtained by a student in the class, we have to find mean.

Correct option is: A) Mean

Correct option is: C) Mean = Median = Mode

For a symmetrical distribution, we have

Mean = Median = Mode

Correct option is: C) Mean = Median = Mode

(3) Arithmetic mean

(C) 11 1/3

Explanation:

We know that,

Average of a data = sum of total observations / total number of observations

According to the question

(x + x+3 + x+5 + x+7 + x+10)/5 = 9

(5x +15)/5 = 9

x + 3 = 9

x = 6

Now, the terms become

6, 6 + 3, 6 + 5, 6 + 7, 6 + 10 = 6, 9, 11, 13, 16

So, the mean of the last three observations = ( 11 + 13 + 16)/3

= 40/3

= 11 1/3

Hence, option (C) is the correct answer.

Correct option is: A) a

Let \(x_1\) 's are original observations.

Let there be n No of observations

\(\therefore\) Sum of observations = \(n\overline x\)

\(\Rightarrow\) \(\sum \limits_{i=1}^n x_i \) = \(n\overline x\) ....(1)

If each observation of data is increased by a then (\(x_i+a\))'s are new observations of data.

Sum of new observations = \(\sum \limits _{i=1}^n (x_i+9) = \sum \limits _{i=1}^n x_i + a \sum \limits _{i=1}^n 1\)

= \(n\overline x\) + na

= n (\(\overline x\) + a)

\(\therefore\) Mean of new observations = \(\frac {sum}n = \frac {n(\overline x + a)}{n} = \overline x + a\)

Hence, if each observation of data is increased by a, then new mean is increases by a.

Correct option is: A) a

A characteristics that varies in magnitude from observation to observation e. g., weight, height, income, age, etc., are variables

The starting and ending values of each class are called lower and upper limit.

(i) \(\frac{S.D.}{mean}\times100\)

(ii) 0, less;

(iii) 11;

(iv) Independent, depends;

(v) Minimum;

(vi) Greater than or equal;

(vii) Square root.

A. equal to that measured from another value

The general formula of Mean is \(\bar x = \frac{\Sigma x_if_i}{\Sigma f_i}\) 

For mean deviation, d = (x-mean)

M.D =  \(\frac{\Sigma fd}{\Sigma f}\)  

If v is the variance and σ is the standard deviation, then 

We know that the formula of standard variance is

\(\sigma = \sqrt{Variance}\) 

So, variance = σ² 

Given,

Mean = 20

\(\frac{\sum yx}{N}=20\)

\(\frac{295+100p+5}{5p+15}=20\)

295 + 100p + 5p² = 100p + 300

295 + 5p² = 300

5p² = 300 – 295

5p² – 5 = 0

5 (p² – 1) = 0

p² – 1 = 0

(p + 1) (p – 1) = 0

p = ± 1

If p + 1 = 0, p = - 1 (Reject)

Or p – 1 = 0, p = 1

False, because the probability of each outcome will be 1/2 only when the two outcomes are equally likely otherwise not.

Graphical representation of a frequency distribution may not be an ogive. It may be a histogram. An ogive is a graphical representation of cumulative frequency distribution.

The mode of the frequency distribution is determined graphically from Histogram.

Answer is (D) 1.1

Answer is (D) 7.5

Answer is (D) 25

Answer is (A) 7

Answer is (B) proportional to the frequencies of the classes.