We are given the frequency distribution of the number of heads when eight coins are tossed 256 times. We are also given that the median number of heads is 4. Question 14 asks for the coefficient of mean deviation about the median. Question 15 asks for the standard deviation of the experiment. Question 16 asks for the geometric mean of the numbers 2, 4, 8, 12, 16, 24. Question 17 asks which measure condenses a large dataset into a single representative value.

1. Problem Description

We are given the frequency distribution of the number of heads when eight coins are tossed 256 times. We are also given that the median number of heads is

4. Question 14 asks for the coefficient of mean deviation about the median.

Question 15 asks for the standard deviation of the experiment.

Question 16 asks for the geometric mean of the numbers 2, 4, 8, 12, 16,

2

4. Question 17 asks which measure condenses a large dataset into a single representative value.

2. Solution Steps

Question 14: Coefficient of Mean Deviation about the Median

The formula for the mean deviation about the median is:

$MD = \frac{\sum f_i |x_i - M|}{N}$

where $f_i$ is the frequency, $x_i$ is the value, $M$ is the median, and $N$ is the total frequency. Here, $N = 256$ and $M = 4$.

We need to calculate $\sum f_i |x_i - 4|$.

$\sum f_i |x_i - 4| = 1|0-4| + 9|1-4| + 26|2-4| + 59|3-4| + 72|4-4| + 52|5-4| + 29|6-4| + 7|7-4| + 1|8-4|$

$= 1(4) + 9(3) + 26(2) + 59(1) + 72(0) + 52(1) + 29(2) + 7(3) + 1(4)$

$= 4 + 27 + 52 + 59 + 0 + 52 + 58 + 21 + 4 = 277$

So, the mean deviation about the median is:

$MD = \frac{277}{256} \approx 1.082$

The coefficient of mean deviation about the median is:

$Coefficient = \frac{MD}{Median} = \frac{1.082}{4} \approx 0.2705$

Since the closest option is (b) 0.222, we must verify the calculation to make sure the correct median was used. Since cumulative frequencies before and including the median are $1+9+26+59+72 = 167$ and the median is 4, this should be correct given $N=256$. It could be the case that if the median is defined as the average of the two middle data points, then the median may be different.

However, it is given that the median is 4, therefore the answer must be close to 0.

2

7

0

5. Therefore there must be a mistake in the problem. The correct answer is close to 0.

2

7

0

5. We will select the closest, but note the issue.

Question 15: Standard Deviation of the Experiment

Let's first compute the mean $\mu = \frac{\sum f_i x_i}{N}$.

$\sum f_i x_i = 1(0) + 9(1) + 26(2) + 59(3) + 72(4) + 52(5) + 29(6) + 7(7) + 1(8)$

$= 0 + 9 + 52 + 177 + 288 + 260 + 174 + 49 + 8 = 1017$

$\mu = \frac{1017}{256} \approx 3.97$

The variance is given by:

$\sigma^2 = \frac{\sum f_i (x_i - \mu)^2}{N}$

$\sigma^2 = \frac{\sum f_i x_i^2}{N} - \mu^2$

$\sum f_i x_i^2 = 1(0^2) + 9(1^2) + 26(2^2) + 59(3^2) + 72(4^2) + 52(5^2) + 29(6^2) + 7(7^2) + 1(8^2)$

$= 0 + 9 + 104 + 531 + 1152 + 1300 + 1044 + 343 + 64 = 4547$

$\sigma^2 = \frac{4547}{256} - (3.97)^2 \approx 17.76 - 15.76 \approx 2.00$

$\sigma = \sqrt{2.00} \approx 1.41$

Question 16: Geometric Mean

The geometric mean of $n$ numbers $x_1, x_2, ..., x_n$ is:

$GM = \sqrt[n]{x_1 x_2 ... x_n}$

For 2, 4, 8, 12, 16, 24, we have $n=6$.

$GM = \sqrt[6]{2 \cdot 4 \cdot 8 \cdot 12 \cdot 16 \cdot 24} = \sqrt[6]{2 \cdot 2^2 \cdot 2^3 \cdot (3 \cdot 2^2) \cdot 2^4 \cdot (3 \cdot 2^3)}$

$GM = \sqrt[6]{2^{1+2+3+2+4+3} \cdot 3^2} = \sqrt[6]{2^{15} \cdot 3^2} = 2^{\frac{15}{6}} \cdot 3^{\frac{2}{6}} = 2^{\frac{5}{2}} \cdot 3^{\frac{1}{3}} = 2^{2.5} \cdot \sqrt[3]{3} = 2^2 \sqrt{2} \sqrt[3]{3} = 4 \cdot 1.414 \cdot 1.442 \approx 8.15$

Question 17: Condensing Data

Averages are measures that condense a dataset into a single representative value.

3. Final Answer

1

4. (b) 0.222 (closest to the actual answer but not entirely correct)

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5. (b) 1.41

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6. (b) 8.2

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7. (d) Averages