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Chapter 6: Problem 15

Find the area under the standard normal curve a. between \(z=0\) and \(z=1.95\) b. between \(z=0\) and \(z=-2.05\) c. between \(z=1.15\) and \(z=2.37\) d. from \(z=-1.53\) to \(z=-2.88\) e. from \(z=-1.67\) to \(z=2.24\)

Short Answer

Expert verified
a) 0.4744 b) 0.4798 c) 0.1162 d) 0.0609 e) 0.9400

Step by step solution

01

Understanding Z-values

Z-values are standard deviations. A positive Z-score tells us how far to the right of the mean our point is, while a negative Z-score tells us how far to the left our point is.
02

Find Area using Z-table (a) between \(z=0\) and \(z=1.95\)

Look up the z-value in the z-table to find the area to the left of that z-value. The z-table gives us 0.9744 (this is to the left of the z-value). Since we want the area from the mean (z=0), we subtract 0.5 (area to the left of z=0) from our looked up area. So, 0.9744 - 0.5 = 0.4744.
03

Find Area using Z-table (b) between \(z=0\) and \(z=-2.05\)

Look up the absolute of z-value (2.05) in the z-table. We get 0.9798, which represents the area to the left. We want the area to the right of -2.05, so we do 1 - 0.9798 = 0.0202. We also subtract this from 0.5 (as we want area from mean), to get 0.5 - 0.0202 = 0.4798.
04

Find Area using Z-table (c) between \(z=1.15\) and \(z=2.37\)

Look up area to left for z=2.37 and z=1.15 in the z-table. We get 0.9911 and 0.8749 respectively. Subtract area corresponding to 1.15 from the area corresponding to 2.37. So, 0.9911 - 0.8749 = 0.1162.
05

Find Area using Z-table (d) from \(z=-1.53\) to \(z=-2.88\)

Look up area to left for absolute z-value of -1.53 and -2.88 in the z-table. We get 0.9370 and 0.9979 respectively. But, we need the areas to the right, so do 1 - 0.9370 = 0.0630 and 1 - 0.9979 = 0.0021.These are the areas to the right of -1.53 and -2.88 respectively. Now subtract area corresponding to 0.0021 from the area corresponding to 0.0630. So, 0.0630 - 0.0021 = 0.0609.
06

Find Area using Z-table (e) from \(z=-1.67\) to \(z=2.24\)

This comprises two parts; the area to the right of -1.67 and the area to the left of 2.24. Look up these two z-values in the z-table to get 0.9525 and 0.9875 respectively. Subtract area to the right of -1.67 from area to the left of 2.24; 0.9875 - (1 - 0.9525) = 0.9400.

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Most popular questions from this chapter

In the Energy Information Administration report The Effect of Income on Appliances in U.S. Households (Source: http://www.eia doe.gov/emeu/recs/appliances/appliances. \(\mathrm{html}\) ), it is noted that \(29 \%\) of housing units with an annual income in the \(\$ 15,000\) to \(\$ 29,999\) range own a large-screen television. Assuming that this is true for the current population of housing units with an annual income in the \(\$ 15,000\) to \(\$ 29,999\) range, find the probability that in a random sample of 400 such housing units, the number that have a large screen television is a. exactly 110 b. 124 to 135 c. no more than 105

One of the cars sold by Walt's car dealership is a very popular subcompact car called Rhino. The final sale price of the basic model of this car varies from customer to customer depending on the negotiating skills and persistence of the customer. Assume that these sale prices of this car are normally distributed with a mean of \(\$ 19,800\) and a standard deviation of \(\$ 350\). a. Dolores paid \(\$ 19,445\) for her Rhino. What percentage of Walt's customers paid less than Dolores for a Rhino? b. Cuthbert paid \(\$ 20,300\) for a Rhino. What percentage of Walt's customers paid more than Cuthbert for a Rhino?

For the standard normal distribution, find the area within \(1.5\) standard deviations of the mean - that is, the area between \(\mu-1.5 \sigma\) and \(\mu+1.5 \sigma\).

The amount of time taken by a bank teller to serve a randomly selected customer has a normal distribution with a mean of 2 minutes and a standard deviation of \(.5\) minute. a. What is the probability that both of two randomly selected customers will take less than I minute each to be served? b. What is the probability that at least one of four randomly selected customers will need more than \(2.25\) minutes to be served?

The highway police in a certain state are using aerial surveillance to control speeding on a highway with a posted speed limit of 55 miles per hour. Police officers watch cars from helicopters above a straight segment of this highway that has large marks painted on the pavement at 1 -mile intervals. After the police officers observe how long a car takes to cover the mile, a computer estimates that car's speed. Assume that the errors of these estimates are nomally distributed with a mean of 0 and a standard deviation of 2 miles per hour. a. The state police chief has directed his officers not to issue a speeding citation unless the aerial unit's estimate of speed is at least 65 miles per hour. What is the probability that a car traveling at 60 miles per hour or slower will be cited for speeding? b. Suppose the chief does not want his officers to cite a car for speeding unless they are \(99 \%\) sure that it is traveling at 60 miles per hour or faster. What is the minimum estimate of speed at which a car should be cited for speeding?
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