A typical incoming telephone call to your catalog sales force results in a mean order of $28.63 with a standard deviation of $13.91. You may assume that orders are received independently of one another. a. Based only on this information, can you find the probability that a single incoming call will result in an order of more than $40? Why or why not? b. An operator is expected to handle 110 incoming calls tomorrow. Find the mean and standard devi- ation of the resulting total order. c. What is the approximate probability distribution of the total order to be received by the operator in part b tomorrow? How do you know? d. Find the (approximate) probability that the operator in part b will generate a total order of more than $3,300 tomorrow. e. Find the (approximate) probability that the operator in part b will generate an average order between $27 and $29 tomorrow.

a. You need to calculate the probability of a single incoming call resulting in an order of more than $40, symbolically P(X>$40). To be able to calculate this probability you need to know what the distribution of the variable is. Without knowing the form of the distribution, i.e. the probability density function, you cannot tell what is the asked probability.

b. If the operator is expected to handle 110 calls in a day (tomorrow) this means that you have a sample of n= 110 calls, and in each call, you are going to take the information of the order placed by the customer.

Note:

If X₁, X₂, ..., Xₙ be the n random variables that constitute a sample, then any function of type θ = î (X₁, X₂, ..., Xₙ) that depends solely on the n variables and does not contain any parameters known, it is called the estimator of the parameter.

When the function i (.) It is applied to the set of the n numerical values of the respective random variables, a numerical value is generated, called parameter estimate θ.

This follows the concepts:

1) The function i (.) It is a function of random variables, so it is also a random variable, that is to say, that every estimator is a random variable.

2) From the above, it follows that Î has its probability distribution and therefore mathematical hope, E (î), and variance, V (î).

And:

The central limit theorem states that if a population with probability function f(X;μ,δ²) from which a random sample of size n is selected. Then the distribution of the sample mean tends to the normal distribution with mean μ and variance δ²/n when the sample size tends to infinity.

As a rule, a sample of size greater than or equal to 30 is considered sufficient to apply the theorem and use the approximation.

This means that if you have the study variable X with a certain distribution, then it's the sample mean X[bar], that is also an aleatory variable, will have the same distribution as it's origin variable. On the other hand, if the distribution of the said variable is unknown, so will be the distribution of its sample mean, but if the sample is large enough, then you can apply the central limit theorem and approximate the distribution of the sample mean to normal, symbolically:

X~?(μ;δ²) and n ≥ 30 then X[bar]≈N(μ;δ²/n )

The mean of said approximation is the same as the mean of the variable of origin.

μ= $28.63

And the standard deviation will be the same as the original variable but affected by the sample size:

δ/√n = $13.91/110= $0.126 ≅ $0.13

c. X[bar]≈N(μ;δ²/n )

d. Using the aproximation you can calculate the asked probabilities with the standard normal:

P(X[bar] > $3,300) = P(Z > [tex]\frac{3.300- 28.63}{13.91/\sqrt{110} }[/tex]) = P(Z > -19.098)= 0

e.

P(27 < X[bat] < 29) = P(X<29) - P(X<27) = P(Z<[tex]\frac{29-28.63}{0.13}[/tex]) - P(Z<[tex]\frac{27-28.63}{0.13}[/tex])

P(Z<0.278) - P(Z<-1.229)= 0.609 - 0.110= 0.499

I hope it helps!