chapter 6

Which of the following is not a characteristic of the normal probability distribution - symmetry - The total area under the curve is always equal to 1 - 99.72% of the time the random variable assumes a value within plus and minus 1 standard deviation of its mean - The mean is equal to the median, which is also equal to the mode - None of the above answers is correct

99.72% of the time the random variable assumes a value within plus and minus 1 standard deviation of its mean

The standard normal distribution

is a normal distribution of standardized values called z-scores.

How is the z-score measured?

it is measured in units of the standard deviation. The z-score tells you how many standard deviations the value x is to the right of the mean or to the left of the mean, μ.

Given that z is a standard normal random variable, find z for each situation.

the area to the left of z is 0.2119 Ans: z = NORM.S.INV(0.2119) = −0.80 the area between −z and z is 0.9030 Ans: z = NORM .S .INV (0.5 + 0.9030/2) = 1.66 the area to the right of z is 0.6915 Ans: z =NORM.S.INV(1−0.6915)=−0.50

The highest point of a normal curve occurs at - one standard deviation to the right of the mean - two standard deviations to the right of the mean - approximately 3 standard deviations to the right of the mean - the mean - None of the above answers is correct

the mean

Finding x values given probabilities

x value with 0.1 in the lower tail =NORM.INV(0.1,B9,B10) x value with 0.25 in the upper tail =NORM.INV(1-0.25,B9,B10)

Converting to Standard Normal Distribution

z = (𝑥−𝜇)/𝜎 We can think of z as a measure of the number of standard deviations x is from 𝜇.

Finding z-values given probabilities

z value with .1 in upper tail = NORM.S.INV(.9) z value with .025 in upper tail = NORM.S.INV(.975) z value with .025 in lower tail = NORM.S.INV(.025)

Standard Normal Probability Distribution

Characteristics - A random variable having a normal distribution with a mean of 0 and a standard deviation of 1 is said to have a standard normal probability distribution. - The letter z is used to designate the standard normal random variable. - Converting to Standard Normal Distribution z = (𝑥−𝜇)/𝜎 We can think of z as a measure of the number of standard deviations x is from mean Excel has two functions for computing probabilities and z values for a standard normal probability distribution. NORM.S.DIST function computes the cumulative probability given a z value. NORM.S.INV function computes the z value given a cumulative probability. "S" in the function names reminds us that these functions relate to the standard normal probability distribution.

Using Excel to Compute Normal Probabilities

Excel has two functions for computing cumulative probabilities and x values for any normal distribution: NORM.DIST is used to compute the cumulative probability given an x value. NORM.INV is used to compute the x value given a cumulative probability.

It was estimated that the mean tire mileage is 36,500 miles with a standard deviation of 5000. The manager now wants to know the probability that the tire mileage x will exceed 40,000. P(x > 40,000) = ?

Solving for the Probability Step 1: Convert x to standard normal distribution. z = (x - mean)/stdev = (40,000 - 36,500)/5,000 = .70 Step 2: Compute the area under the standard normal curve to the right of z = .7 P(z > .70) = 1 - P(z < .70) = 1- NORM.S.DIST(.70,TRUE) = .2420

Normal Probability Distribution

Characteristics - The distribution is symmetric; its skewness measure is zero. - The entire family of normal probability distributions is defined by its mean m and its standard deviation s . - The highest point on the normal curve is at the mean, which is also the median and mode. - The mean can be any numerical value: negative, zero, or positive. - Area under curve =1 - The standard deviation determines the width of the curve: larger values result in wider, flatter curves. - Probabilities for the normal random variable are given by areas under the curve. The total area under the curve is 1 (.5 to the left of the mean and .5 to the right). - (basis for the empirical rule) 68.3% of values of a normal random variable are within +/- 1 standard deviation of its mean. 95.4% of values of a normal random variable are within +/- 2 standard deviations of its mean. 99.7% of values of a normal random variable are within +/- 3 standard deviations of its mean.

What does the z-score allow us to do?

It allows us to compare data that are normally distributed but scaled differently.

If the mean of a normal distribution is negative, - the standard deviation must also be negative - the variance must also be negative - a mistake has been made in the computations, because the mean of a normal distribution can not be negative - the standard deviation must be 0 - None of the above answers is correct

None of the above answers is correct The mean can be negative, or positive. Think about if you are looking at temperatures in the North Below where they are almost always below 0, which means the average temperature is going to be negative. The variance and standard deviation is always positive.

Probabilities Normal Distribution

P(X<=20,000) =NORM.DIST(20000,MEAN,STDDEV,TRUE) P(20000<=x<=40000) = =NORM.DIST(40000,B9,B10,TRUE)-NORM.DIST(20000,B9,B10,TRUE) P(x>=40000) = =1-NORM.DIST(40000,B9,B10,TRUE)

Excel Probabilities: Standard Normal Distribution

P(Z<=1) = NORM.S.DIST(1,TRUE) P(-.5<=Z<=1.25) = NORM.S.DIST(1.25,TRUE)-NORM.S.DIST(-.5,TRUE) P(Z>=1.58) = 1-NORM.S.DIST(1.58,TRUE)

Given that z is a standard normal random variable, compute the following probabilities.

P(z < 1.20) = NORM.S.DIST(1.20,TRUE) = 0.8849 P(z ≥ −0.23) = 1 − NORM .S .DIST (−0.23, TRUE ) = 0.5910 P(−1.98 ≤ z ≤ 0.49) = NORM .S .DIST (0.49, TRUE ) − NORM.S.DIST(−1.98,TRUE) = 0.6641

What should be the guaranteed mileage if Grear wants no more than 10% of tires to be eligible for the discount guarantee? (Hint: Given a probability, we can use the standard normal table in an inverse fashion to find the corresponding z value.)

Solving for the guaranteed mileage Step 1: Find the z value that cuts off an area of .1 in the left tail of the standard normal distribution. z=NORM.S.INV(0.1)=-1.28 We see that z = -1.28 cuts off an area of 0.1 in the lower tail. Step 2: Convert z to the corresponding value of x. x = mean + z (stdev) x = 36500 - 1.28 (5000) = 30092.24 Thus a guarantee of 30092.24 miles will meet the requirement that approximately 10% of the tires will be eligible for the guarantee.

During early 2012, economic hardship was stretching the limits of France's welfare system. One indicator of the level of hardship was the increase in the number of people bringing items to a Paris pawnbroker. That number had risen to 658 per day. Assume the number of people bringing items to the pawnshop per day in 2012 is normally distributed with a mean of 658.

Suppose you learn that on 3% of the days, 610 or fewer people brought items to the pawnshop. What is the standard deviation of the number of people bringing items to the pawnshop per day? Ans: NORM .S .INV (0.03) = 610−658 / σ σ = 610−658 / NORM .S .INV (0.03) = 25.52 On any given day, what is the probability that between 600 and 700 people bring items to the pawn shop? Ans: P (600 < x < 700) = NORM .DIST (700, 658, 25.52, TRUE ) − NORM .DIST (600, 658, 25.52, TRUE ) = 0.9386 What is the cutoff count value for the busiest 3% of days? Ans: NORM.INV(1−0.03,658,25.52)=706.00

The average price for a gallon in the United States is $3.73 and in Russia it is $3.40. Assume these averages are the population means in the two countries and that the probability distributions are normally distributed with a standard deviation of $0.25 in the United States and a standard deviation of $0.20 in Russia.

What is the probability that a randomly selected gas station in the United States charges less than $3.50 per gallon? P(X<=3.5) Ans: NORM.DIST(3.50,3.73,0.25,TRUE)=0.1788 What percentage of gas stations in Russia charge more than $3.50 per gallon? P(x>=3.5) Ans: 1−NORM.DIST(3.50,3.40,0.20,TRUE)=0.3085 What is the minimum gas price for the top 10% in the United States? z value with .1 in upper tail = NORM.S.INV(.9) Ans: NORM.INV(0.9,3.73,0.25)=4.05 What is the cutoff value for the lowest 10% gas prices in the United States? Ans: NORM.INV(0.1,3.73,0.25)=3.41