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Chapter 6: Problem 86

Calculate the area under the standard normal curve to the left of these values: a. \(z=-.90\) b. \(z=2.34\) c. \(z=5.4\)

Short Answer

Expert verified
Answer: a. 18.41%, b. 99.06%, and c. approximately 100%.

Step by step solution

01

Refer to the standard normal table or use a calculator/software

Find a standard normal table (also called the \(z\)-table) or use a calculator or software that provides the area to the left of the given \(z\)-score. Remember, the value from the table or calculator will give you the probability.
02

Identify the \(z\)-score and look up or calculate the area to the left

For each of the given \(z\)-scores, locate the value in the \(z\)-table or calculate the area to the left of the particular value using a calculator or software. a. \(z=-.90\) Using a standard normal table, the area to the left of the \(z\)-score \(-0.90\) is 0.1841. b. \(z=2.34\) Look up the \(z\)-score \(2.34\) in the \(z\)-table or calculate the area to the left using calculator/software; the value is 0.9906. c. \(z=5.4\) The \(z\)-table may not display values as high as \(5.4\). However, you could use a calculator or software to find the area to the left of the \(z\)-score \(5.4\). The value is approximately 1.0000 since the area under the curve is essentially the entire distribution at this point.
03

Interpret the results

Now that we have the area under the curve to the left of each \(z\)-score, we can interpret the results: a. For \(z=-0.90\), there is an 18.41% (0.1841) chance that a value from the standard normal distribution will be less than or equal to \(-0.90\). b. For \(z=2.34\), there is a 99.06% (0.9906) chance that a value from the standard normal distribution will be less than or equal to \(2.34\). c. For \(z=5.4\), there is essentially a 100% (1.0000) chance that a value from the standard normal distribution will be less than or equal to \(5.4\).

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Most popular questions from this chapter

How does the IRS decide on the percentage of income tax returns to audit for each state? Suppose they do it by randomly selecting 50 values from a normal distribution with a mean equal to \(1.55 \%\) and a standard deviation equal to \(.45 \% .\) (Computer programs are available for this type of sampling.) a. What is the probability that a particular state will have more than \(2.5 \%\) of its income tax returns audited? b. What is the probability that a state will have less than \(1 \%\) of its income tax returns audited?

Human heights are one of many biological random variables that can be modeled by the normal distribution. Assume the heights of men have a mean of 69 inches with a standard deviation of 3.5 inches. a. What proportion of all men will be taller than \(6^{\prime} 0^{\prime \prime}\) ? (HINT: Convert the measurements to inches.) b. What is the probability that a randomly selected man will be between \(5^{\prime} 8^{\prime \prime}\) and \(6^{\prime} 1^{\prime \prime}\) tall? c. President George \(\mathrm{W}\). Bush is \(5^{\prime} 11^{\prime \prime}\) tall. Is this an unusual height? d. Of the 42 presidents elected from 1789 through 2006,18 were \(6^{\prime} 0^{\prime \prime}\) or taller. \(^{1}\) Would you consider this to be unusual, given the proportion found in part a?

The number of times \(x\) an adult human breathes per minute when at rest depends on the age of the human and varies greatly from person to person. Suppose the probability distribution for \(x\) is approximately normal, with the mean equal to 16 and the standard deviation equal to \(4 .\) If a person is selected at random and the number \(x\) of breaths per minute while at rest is recorded, what is the probability that \(x\) will exceed \(22 ?\)

A researcher notes that senior corporation executives are not very accurate forecasters of their own annual earnings. He states that his studies of a large number of company executive forecasts "showed that the average estimate missed the mark by \(15 \%\)." a. Suppose the distribution of these forecast errors has a mean of \(15 \%\) and a standard deviation of \(10 \%\). Is it likely that the distribution of forecast errors is approximately normal? b. Suppose the probability is .5 that a corporate executive's forecast error exceeds \(15 \% .\) If you were to sample the forecasts of 100 corporate executives, what is the probability that more than 60 would be in error by more than \(15 \% ?\)

Calculate the area under the standard normal curve between these values: a. \(z=-1.4\) and \(z=1.4\) b. \(z=-3.0\) and \(z=3.0\)
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