Question

A chemical supply company currently has in stock 100 pounds of a certain chemical, which it sells to customers in 5 -pound lots. Let \(x=\) the number of lots ordered by a randomly chosen customer. The probability distribution of \(x\) is as follows: $$ \begin{array}{lcccc} x & 1 & 2 & 3 & 4 \\ p(x) & 0.2 & 0.4 & 0.3 & 0.1 \end{array} $$ a. Calculate and interpret the mean value of \(x\). b. Calculate and interpret the variance and standard deviation of \(x\).

Short answer

The mean value of \(x\) is 2.3. This can be interpreted as the expected value or the average amount of 5-pound lots ordered by a customer. The variance and the standard deviation are 0.61 and approximately 0.78, respectively, indicating a relatively small spread around the mean.

Step-by-step solution

Calculate the Mean

To compute the average, use the formula for the mean of a discrete probability distribution, which is \(\mu = \Sigma (x \cdot p(x))\). In other words, each possible outcome should be multiplied by its associated probability and then all these products should be summed. Using the given distribution values: \(\mu = (1*0.2) + (2*0.4) + (3*0.3) + (4*0.1) = 2.3\). The mean value of \(x\) is 2.3.

Calculate the Variance

Next, calculate variance (\(\sigma^2\)) from the formula for the variance of a discrete probability distribution, \(\sigma^2 = \Sigma ((x- \mu)^2 \cdot p(x))\). For each potential outcome \(x\), subtract the mean (\(\mu\)) and square the result, multiply this by the corresponding probability, finally sum all these products beyond all distinct values of \(x\): \(\sigma^2 = ((1-2.3)^2*0.2) + ((2-2.3)^2*0.4) + ((3-2.3)^2*0.3) + ((4-2.3)^2*0.1) = 0.61\). So, the variance is 0.61.

Calculate the Standard Deviation

For the final step, obtain the standard deviation (\(\sigma\)). The standard deviation is just the square root of the variance. Using the variance from Step 2, take the square root: \(\sigma = \sqrt{0.61} \approx 0.78\). The standard deviation is approximately 0.78.