ECON 245: Descriptive Statistics and Probability

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ECON 245: Descriptive Statistics and Probability

Laboratory No. 11: Continuous Distributions
Instructions: Submit all your output by compiling a LATEXdocument. Ensure that your floats are correct, for where you place the graphs or tables.
Source: Appendix 6.2.

Objective: practising using continuous probability distributions with Excel.

Continuous Distributions

Excel function for computing cumulative probabilities for a normal distribution is NORM.D IST. It has four arguments:
• x
• µ, mean
• σ, standard deviation
• and cumulative.
– TRUE is used if we want the cumulative probability.

E.g.: Sup. a Tire Company developed a new tire to be sold in their stores. Since it is a new product, the managers believe that offering a mileage guarantee will be important for customers to accept the product. Thus, to finalize the tire mileage guarantee, the managers want probability information about x = the number of miles the tires will last.

The engineers believe:
• µ = 36, 500.
• σ = 5, 000.
• It is reasonable to assume a normal distribution.
Sup. we are interested in the probability that tire mileage exceeds 40,000 miles.

The function NORM.D IST can be used to compute cumulative probabilities for a normal distribution.

To compute a cumulative probability that the tire mileage will be less than or equal to 40,00 miles we need to enter the following formula into any cell:

= NORM.D IST(40000, 36500, 5000, TRUE).
- The number that appears in the cell (0.758) will be the probability of tire mileage being less than or equal to 40,000.
• Thus, the probability that tire mileage will exceed 40,00 miles is:
1 − 0.758 = 0.2420

Note: TRUE can also be imputed as the number 1. E.g.: = NORM.D IST(40000, 36500, 5000, 1).

Excel also has the NORM.INV function which uses an inverse computation to find the x value corresponding to a given cumulative probability.

For instance, sup. we are interested instead in finding the guaranteed mileage that should be offered so that no more than 10% of the tires are eligible for the guarantee, then:

= NORM.INV(.1, 36500, 5000).
- The number that appears in the cell (30,092) will be the probability of tire lasting 30,092 or less is 10%

Excel’s NORM.S.D IST can be used to compute probabilities for r.v./ that follow a standard normal distribution. The general form of the function is: NORM.S > D IST(z, cumulative) where, z refers to the z value corresponding to the random variable and TRUE is specified for the second argument if a cumulative probability is what we are after. Similarly, NORM.S.INV function uses an inverse computation to find the z value corresponding to a given cumulative probability. The general form of the function is: NORM.S.INV(probability) where the only argument needed is the cumulative probability to be used. The function returns the corresponding z value.

Lastly, the function for computing exponential probabilities is EXPON.D IST, which requires three inputs:

• x, the value fo the variable;
• lambda, which is 1/µ;
• TRUE, if computing the cumulative probability.

E.g.: an exponential probability distribution with mean, µ = 15. The probability that the exponential variable is less

than or equal to 6 can be computed by:
= EXPON.D IST(6, 1/15, TRUE).
- The number that appears in the cell (0.3297) indicates that the probability the exponential variable will be less than or equal to 6 is 0.3297.
1. z is a standard normal random variable. The P(1.5 ≤ z ≤ 2.3) is?
2. z is a standard normal random variable. P(z ≥ 2.11) equals?
3. Scores on a recent national statistics exam were normally distributed with a mean of 80 and a standard deviation of 8.
a. What is the probability that a randomly selected exam will have a score of at least 71?
b. What percentage of exams will have scores between 89 and 92?
c. If the top 2.5% of test scores receive merit awards, what is the lowest score eligible for an award?
4. . The average starting salary of this year’s MBA students is $45,000 with a standard deviation of $5,000. Furthermore, it is known that the starting salaries are normally distributed. What are the minimum and the maximum starting salaries of the middle 95% of MBA graduates?
5. The average starting salary for this year’s graduates at a large university (LU) is $35,000 with a standard deviation of $6,000. Furthermore, it is known that the starting salaries are normally distributed.
a. What is the probability that a randomly selected LU graduate will have a starting salary of at least $38,400?
b. Individuals with starting salaries of less than $19,600 receive a low income tax break. What percentage of the graduates will receive the tax break?
c. What are the minimum and the maximum starting salaries of the middle 98% of the LU graduates?

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