The Normal DistributionNotes and review problems

The Normal DistributionA normal variable is continuous, ranges from to +, and forms a symmetrical distribution, with a mean m and a standard deviation sm

The Normal DistributionThe density function f(x) is m

Solve the following problems1.Calculate the following probabilities:P(0

SolutionsSet 1:P(0

SolutionsSet 2:P(Z > Z0) = .04. Thus, P(Z < Z0) = 0.96. The corresponding Z0= 1.75 With Excel use: =normsinv(.96)P(Z < Z0) = .02. The value of Z0 must be negative, therefore, if we want to use the text table we need to use symmetry. Instead of finding Z0 as requested, well first find Z0, which of course has a positive value! For convenience lets call this value Z1 P(Z > Z1) = .02. So, P(Z < Z1) = 0.98. Thus Z1 = 2.055 (or so). This means that the value we are seeking Z0 = -2.055.P(-Z0 < Z < Z0) = 0.65. Then P(Z < Z0) P(Z Z0) = 1 P(Z < Z0), we have: P(-Z0 < Z < Z0) = P(Z < Z0) [1 P(Z < Z0)] = 2P(Z < Z0) 1 = 0.65. From here we have P(Z < Z0) = 0.825, which yields Z0 =~0.935. Another solution: Note that P(ZZ0)=1, but P(ZZ0) so, P(Z

SolutionsSet 3:P(X>145) = P(Z>(145 100)/20) = P(Z>2.25). Use the Z-table. With Excel use: =1 normdist(145,100,20,True)P(X

4.The lifetime of light-bulbs that are advertised to last for 5000 hours are normally distributed with a mean of 5,100 hours and a standard deviation of 200 hours.What is the probability that a bulb lasts longer than the advertised figure?If we wanted to be sure 98% of all the bulbs last longer than an advertised figure, what figure should be advertised? Because of high interest rates, most consumers attempt to pay-off the credit card bills promptly. However this is not always possible. If the amount of interest paid monthly by card-holders is normally distributed with a mean of $27 and a standard deviation of $7,What proportion of the cardholders pay more than $30 in interest?What proportion of the cardholders pay less than $15 in interest?What interest payment is exceeded by only 20% of the cardholders?

Solve the following problems

Solutions4.Let X represent the bulb lifetime.P(X>5100) = P(Z>(5000 5100)/200) = P(Z>-0.5) = P(ZX0) = .98; X0 must reside to the left of m, therefore after transforming X into a Z variable, the latter must have a negative value. Let us call this Z-value Z0 where Z0 = (X0 m)/s. Thus P(Z> Z0) = .98 (click to see the graph), so P(Z

Solutions5.Let X be the amount of interest paid monthly.P(X>30) = P(Z>(30-27)/7) = P(Z>.42857). With Excel we have: = 1-normdist(30,27,7,true) =.334118P(X

Sampling DistributionsNotes and review problems

Sampling Distribution of the Sample MeanThe sample mean is normally distributed if the parent distribution is normal.Then sample mean is approximately normally distributed if the sample is sufficiently large (n30) even if the parent distribution is not normal. The larger the sample the better the approximation.The mean of the sample mean is the same as the parent population meanThe standard deviation of the sample mean is

Solve the following problemsA sample of n=16 observations is drawn from a normal population with m=1,000 and s=200. Find the following probabilities: The heights of North American women are normally distributed with a mean of 64 inches and a standard deviation of 2 inches.What is the probability that a randomly selected woman is taller than 66 inches?A random sample of 40 women is selected. What is the probability that the sample mean height is greater than 66 inches?If the population of heights is not normally distributed, would you change your answer to part 2?The manufacturer of cans of salmon claims the average can weight is 6.05 ounces. He further states the actual weight per can is a normal variable with a mean of 6.05 and standard deviation of 0.18 ounces. A random sample of 36 observations is drawn.What is the probability the sample mean weight is less 5.97?If the random sample indeed produced a mean weight less than 5.97. Comment on the statement made by the manufacturer.

SolutionsProblem 1 Problem 2 Let X be a woman height. Problem 3

The manufacturer is probably wrong. The chance to have a sample mean as small as 5.97 is extremely small if the mean is 6.05. So if we actually have this sample mean, the real population mean must be smaller.

Solve the following problems4.A portfolio can be composed of either one of two stocks, or of both. The mean and standard deviation of the rate of return for each stock are .12 and .05. The returns are normally distributedIf an investor decides to invest in one stock only, what is the probability that the portfolio loses money? Earns more than 20%?if the investor decides to invest in both stocks, one dollar in each stock, and assuming the return on each stock is independent on the otherwhat is the probability that the portfolio loses money? what is the probability that the portfolio earns more than 20%? what do the above results tell you about diversification (both advantage and disadvantage)?Now assume the return and standard deviation for stock 2 are .24 and .15 respectively. Answer again parts a and b. For part b the portfolio return and standard deviation depend on the proportion of the portfolio invested in each stock. Assume for simplicity that the ratio is as before - a dollar for dollar. Try to resolve part c when the ratio of investments in the two stocks for the portfolio is 2 dollars in stock 1 for 1 dollar in stock 2. Note that the return in this case is .67R1+.33R2 for each dollars invested in the portfolio. From this relationship you can determine the mean and standard deviation of the portfolio.

Solutions4.Let X be the rate of return for each stockP(X1.6) = Let X1 and X2 be the return for each stock. The portfolio loses money when X1 + X2 < 0. To calculate the probability P(X1 + X2

5.The signs on an elevator states Maximum Capacity 1,140 Kilogram or 16 persons. If the mean weight is 75 kilos and the standard deviation of the weight is 10 kilos,What is the probability that a sample of 16 persons would weigh more than 1140 kilos.What is the probability that 18 individuals will have a weight of more than 80 Kilos out of this sample of 36 individuals?

Solve the problems

Solutions5.a. P(X1+X2++X16>1140)= b.We use the normal distribution to find P(X>80). Then we use the binomial distribution to calculate P(Binomial = 18), where a success is defined as the individual weight is more than 80 pounds. P(X>80) = P(Z>(80 75)/10) = P(Z>.5) = .3085. This is the probability of a success for the binomial distribution we are about to use. Define B as a binomial variable with n = 36 and p = .3085. P(B=18) = binomdist(18,36,.3085,false)= 0.00762. If you are unfamiliar with the function binomdist in Excel you can use the normal approximation to the binomial (since youll probably not find a binomial table for n=36). The mean of this normal distribution is m =np=36(.3085)=11.12, and the standard deviation is s =[np(1-p)]1/2=[(36)(.3085)(1-.3085)]1/2=2.77. The probability to calculate is P(B=18)@P(17.5X18.5), where X is the normal variable with the pre-calculated m and s. The result is .006715

Sampling Distribution of the Sample Proportion (large samples)If x represents a count of the number of successes (the number of occurrences of a certain event) then p = x/n represents the proportion of successes in the sample.The sample proportion is approximately normally distributed with the following parameters:The mean is the real proportion in the population: mp = pThe standard deviation is:

Solve the problems1.A manufacturer of aspirin claims that the proportion of headaches sufferers who get relief with just two aspirins is 53%.What is the probability in a random sample of 400 headache sufferers, less than 50% obtain relief?If indeed 50% of the sample actually obtained relief, what does this suggest about the manufacturers claim?2.A university bookstore claims that 50% of its customers are satisfied with the service and priceIf the claim is true, what is the probability that in a random sample of 600 customers, less than 45% are satisfied?If 270 customers expressed satisfaction with the bookstore, what does this tell you about the bookstores claim?

SolutionsThe probability to calculate is P(sample proportion 5; np(1 p) = 400(1-.53) = 188> 5. Approximating the distribution of the sample proportion by the normal distribution should lead to accurate results. By the manufacturers statement: m= .53. s=[p(1-p)/n]1/2 = [.53(1-.53)]1/2= .024955. P(Proportion 5. P(Proportion