Question

What is the mean, variance and standard deviation of the sum of a $4$ sided die and an $8$ sided die? Use these values to estimate the mean, variance and standard deviation of the sum of $30$, $4$sided dice plus $30$, $8$sided dice. Plot the probability distribution of the estimated sum. What is the approximate probability of the sum being greater than $150$? You may want to use graph paper for this question.

Answer 1

Statistics is the study of the collection, analysis, interpretation, presentation, and organization of data. In the given question we have to calculate the mean, the standard deviation, and the variance. Mean is the average of a given set of data. Variance is the sum of differences between all numbers and means. Standard deviation is a measure of the extent to which data varies from the mean.

Complete step-by-step answer:
In the above question we have to calculate the mean, variance and standard deviation. We have formulas to calculate all these values.
Mean is the average of the observation. The formula for mean is:
$\Rightarrow mean=\frac{\sum{x}}{n}$
Variance is the measure of spread out of the collection of data. The formula to find the variance is:
\Rightarrow {{\sigma }^{2}}=\frac{\sum{{{\left( {{x}_{i}}-\overset{\\_}{\mathop{x}}\, \right)}^{2}}}}{N}
Standard deviation is the measure of the dispersion of data from the mean. The square of standard deviation is equal to the variance. The formula to find the standard deviation is:
\Rightarrow \sigma =\sqrt{\frac{\sum{{{\left( {{x}_{i}}-\overset{\\_}{\mathop{x}}\, \right)}^{2}}}}{N}}
Now, the possible outcomes of throwing the $4$ sided die are $1,2,3$ and$4$.
So the mean is:
\begin{aligned} & \Rightarrow \overset{\\_}{\mathop{x}}\,=\frac{1+2+3+4}{4} \\\ & \Rightarrow \overset{\\_}{\mathop{x}}\,=\frac{10}{4}=2.5 \\\ \end{aligned}
Now we know that the variance is equal to:
$\begin{aligned} & \Rightarrow {{\sigma }^{2}}=\frac{{{\left( 1-2.5 \right)}^{2}}+{{\left( 2-2.5 \right)}^{2}}+{{\left( 3-2.5 \right)}^{2}}+{{\left( 4-2.5 \right)}^{2}}}{4} \\\ & \Rightarrow {{\sigma }^{2}}=\frac{2.25+0.25+0.25+2.25}{4} \\\ & \Rightarrow {{\sigma }^{2}}=\frac{5}{4}=1.25 \\\ \end{aligned}$
The possible outcomes of throwing the $8$ sided die are $1,2,3,4,5,6,7,8$.
Now the mean is:
\begin{aligned} & \Rightarrow \overset{\\_}{\mathop{x}}\,=\frac{1+2+3+4+5+6+7+8}{8} \\\ & \Rightarrow \overset{\\_}{\mathop{x}}\,=4.5 \\\ \end{aligned}
Now the variance is:
$\begin{aligned} & \Rightarrow {{\sigma }^{2}}=\frac{{{\left( 1-4.5 \right)}^{2}}+{{\left( 2-4.5 \right)}^{2}}+{{\left( 3-4.5 \right)}^{2}}+{{\left( 4-4.5 \right)}^{2}}+{{\left( 5-4.5 \right)}^{2}}+{{\left( 6-4.5 \right)}^{2}}+{{\left( 7-4.5 \right)}^{2}}+{{\left( 8-4.5 \right)}^{2}}}{8} \\\ & \Rightarrow {{\sigma }^{2}}=\frac{12.25+6.25+2.25+0.25+0.25+2.25+6.25+12.25}{8} \\\ & \Rightarrow {{\sigma }^{2}}=5.25 \\\ \end{aligned}$
Now the mean of the sum of the two dice is the sum of the means, so we get
$\begin{aligned} & \Rightarrow 2.5+4.5 \\\ & \Rightarrow 7 \\\ \end{aligned}$
And the variance is also the sum of the two variances:
$\begin{aligned} & \Rightarrow 1.25+5.25 \\\ & \Rightarrow 6.5 \\\ \end{aligned}$
As we discussed above the standard deviation is just the square root of the variance, so the standard deviation is:
$\Rightarrow \sqrt{6.5}=2.549$
If we have $30$ $4$ sided dice and $30$ $8$ sided dice, we get:
\begin{aligned} & \Rightarrow \overset{\\_}{\mathop{x}}\,=7\times 30=210 \\\ & \Rightarrow {{\sigma }^{2}}=6.5\times 30=195 \\\ & \Rightarrow \sigma =\sqrt{195}=13.964 \\\ \end{aligned}
The estimated sum will be approximately normally distributed with mean$210$ and standard deviation is $13.964$:
$N\left[ 210,13.964 \right]$.
Now we have to calculate $P\left[ sum>150 \right]$ . for this we go to the normalized normal distribution, then we get
$\begin{aligned} & \Rightarrow z=\frac{\left( 149.5-210 \right)}{13.964} \\\ & \Rightarrow z=-4.3325 \\\ \end{aligned}$
Here in the above expression instead of $150$ we used$149.5$ because of continuity correction.
If we look at the value of z in the table we get the value of $P\left[ sum>150 \right]$is $0.9999...$ .
Hence by using a table we get the value of $P\left[ sum>150 \right]$which is $0.9999...$.

Note: In the table of the value of z we have to find out the least value of the z. otherwise our answer may be affected. Here the word which we use is normal distribution , it is nothing; it is a continuous probability distribution that is symmetrical on both sides of the mean, so the right side of the centre is a mirror image of the left side.