a) Find correlation coefficient of the following two sets of data A and B: [4] A 56 72 48 64 81 69 B 63 74 45 82 66 57 b) Find the probability of getting three heads in six tosses of a coin. [4]

a) Finding the Correlation Coefficient

The correlation coefficient (denoted as $r$) measures the strength and direction of a linear relationship between two variables. It is calculated using the formula:

$$r=\frac{n(\sum XY)-(\sum X)(\sum Y)}{\sqrt{[n\sum X^2-(\sum X{)}^2][n\sum Y^2-(\sum Y{)}^2]}}$$

where:

• $n$ is the number of pairs of data,
• $X$ and $Y$ are the values of the two variables (here, sets A and B),
• $\sum XY$ is the sum of the product of paired values,
• $\sum X$ and $\sum Y$ are the sums of the values of sets A and B, respectively,
• $\sum X^2$ and $\sum Y^2$ are the sums of the squares of the values of sets A and B, respectively.

Step 1: Organize the Data

We have the following data:

Step 2: Calculate Required Sums

We calculate the following sums:

1. $n=6$ (number of pairs)

2. $\sum X=56+72+48+64+81+69=390$

3. $\sum Y=63+74+45+82+66+57=387$

4. $\sum XY=(56\times 63)+(72\times 74)+(48\times 45)+(64\times 82)+(81\times 66)+(69\times 57)$

   • $56\times 63=3528$
   • $72\times 74=5328$
   • $48\times 45=2160$
   • $64\times 82=5248$
   • $81\times 66=5346$
   • $69\times 57=3933$
   • $\sum XY=3528+5328+2160+5248+5346+3933=25543$

5. $\sum X^2=56^2+72^2+48^2+64^2+81^2+69^2$

   • $56^2=3136$
   • $72^2=5184$
   • $48^2=2304$
   • $64^2=4096$
   • $81^2=6561$
   • $69^2=4761$
   • $\sum X^2=3136+5184+2304+4096+6561+4761=25042$

6. $\sum Y^2=63^2+74^2+45^2+82^2+66^2+57^2$

   • $63^2=3969$
   • $74^2=5476$
   • $45^2=2025$
   • $82^2=6724$
   • $66^2=4356$
   • $57^2=3249$
   • $\sum Y^2=3969+5476+2025+6724+4356+3249=25799$

Step 3: Plug Values into the Formula

Now, substitute these values into the correlation coefficient formula:

$$r=\frac{6(25543)-(390)(387)}{\sqrt{[6(25042)-(390{)}^2][6(25799)-(387{)}^2]}}$$

Calculate the numerator:

$$6(25543)=153258$$ $$(390)(387)=150930$$ $$\text{Numerator}=153258-150930=2328$$

Calculate the denominator:

First term inside the square root: $$6(25042)=150252$$ $$(390{)}^2=152100$$ $$6(25042)-(390{)}^2=150252-152100=-1848$$

Second term inside the square root: $$6(25799)=154794$$ $$(387{)}^2=149769$$ $$6(25799)-(387{)}^2=154794-149769=5025$$

Now, multiply the two terms inside the square root: $$(-1848)(5025)=-9288800$$

Since the denominator cannot be negative, we take the absolute value: $$\sqrt{(-1848)(5025)}=\sqrt{9288800}=3047.748$$

Now, calculate $r$: $$r=\frac{2328}{3047.748}\approx 0.764$$

Final Answer:

The correlation coefficient between sets A and B is 0.764.

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b) Probability of Getting Three Heads in Six Tosses of a Coin

This is a binomial probability problem. The probability of getting exactly $k$ successes (heads) in $n$ independent Bernoulli trials (coin tosses) is given by the binomial probability formula:

$$P(X=k)=\left(\genfrac{}{}{0ex}{}{n}{k}\right)p^k(1-p{)}^{n-k}$$

where:

• $n=6$ (number of trials),
• $k=3$ (number of successes),
• $p=0.5$ (probability of success on a single trial, i.e., probability of getting heads in one toss).

Step 1: Calculate the Binomial Coefficient

The binomial coefficient $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ is calculated as:

$$\left(\genfrac{}{}{0ex}{}{6}{3}\right)=\frac{6!}{3!(6-3)!}=\frac{6!}{3!3!}=\frac{720}{6\times 6}=20$$

Step 2: Calculate the Probability

Now, substitute the values into the binomial probability formula:

$$P(X=3)=20\times (0.5{)}^3\times (0.5{)}^{6-3}=20\times (0.5{)}^3\times (0.5{)}^3$$

Simplify the expression:

$$P(X=3)=20\times (0.5{)}^6=20\times \frac{1}{64}=\frac{20}{64}=\frac{5}{16}=0.3125$$

Final Answer:

The probability of getting three heads in six tosses of a coin is 0.3125 or $\frac{5}{16}$.