Advanced Math Topics Chapters 8 and 9 Review. The average purchase by a customer in a large novelty store is $4.00 with a standard deviation of $0.85.

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Presentation transcript:

Advanced Math Topics Chapters 8 and 9 Review

The average purchase by a customer in a large novelty store is $4.00 with a standard deviation of $0.85. If 49 customers are selected at random, what is the probability that their average purchase will be less than $3.70? μ x = $4.00 x = $3.70 z = Look this up in the columns in the chart – =.0068 = 0.68%

Remember this from last chapter? How would you find the limits of the sample means that lie in the middle 95% of the normal distribution? μxμx.475 Lower LimitUpper Limit If you did not know the mean of a population (μ), but you knew a sample mean ( x ), it would help to have a formula for μ. Rearrange the formulas above to get…

A sample survey of 81 movie theaters showed that the average length of the main feature film was 90 minutes with a standard deviation of 20 minutes. Find a… 95% confidence interval for the mean of the population. The 95% confidence interval for μ is to minutes.

Taking a sample can be a time consuming and costly event. Therefore, it helps to have a formula to figure out how big a sample we need. e = the maximum allowable error n = sample size Then, a sample of size… …will result in an estimate of μ within the maximum allowable error 95% of the time. What happens to the sample size as (e) gets larger? Does this make sense?

The management of a large company in California wants to estimate the average working experience in years of its 2000 hourly workers. How large a sample should be taken in order to be 95% confident that the sample mean does not differ from the population mean by more than ½ half of a year? Past experience indicates that the standard deviation is 2.6 years. n = The sample should consist of at least 104 workers.

It is known that 70% of all airplane tickets sold by a certain company are round-trip tickets. A random sample of 100 passengers is taken. What is the probability that at least 75% of these passengers have round trip tickets? There is a large population in which a proportion of the population has a certain characteristic. We select random samples of size n and determine the proportion in each sample with the characteristic. Then the sample proportions will be approximately normally distributed with mean p (the population proportion) and standard deviation… p = 70% p = 75% z = – = = 13.79% Similar to our z formula…

HW P. 432 #2,4,15,16 P. 470 #2, 5