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Chapter 6: Problem 31

Determine the area under a normal distribution curve with \(\mu=55\) and \(\sigma=7\) A. to the right of \(x=58\) b. to the right of \(x=43\) c. to the left of \(x=68\) d. to the left of \(x=22\)

Short Answer

Expert verified
To get the short answers, after calculating the Z-scores, the values are looked up in the Z-table. For cases (a) and (b), subtract this value from 1. For cases (c) and (d), use the value directly. These provide the areas under the normal distribution curve for each scenario.

Step by step solution

01

Calculate the Z-scores

The Z-score for a value \(x\) from a distribution with mean \(\mu\) and standard deviation \(\sigma\) is given by the formula \( Z = (x-\mu)/\sigma \). Thus, we calculate the four Z-scores: Z1 for \(x = 58\), Z2 for \(x = 43\), Z3 for \(x = 68\), and Z4 for \(x = 22\). Note that the Z-score will indicate how many standard deviation units \(x\) is from the mean.
02

Use the Z-table

Next, we move to finding the area relevant to our Z-scores from the Z-table. The Z-table helps us find the probability that a statistic is less than our observed Z-score.
03

Adjust based on the scenario

In cases (a) and (b), we're asked for the area 'to the right'. However, the standard Z-table gives the area 'to the left'. For scenarios (a) and (b), we subtract the probabilities we find from 1 since the total area under the curve is 1.
04

Calculation for each case

For case (a), find the value in the Z-table corresponding to Z1 and subtract this from 1. Repeat similarly for case (b). In cases (c) and (d), use the Z-table to find the corresponding probabilities as it is for 'left of' scenarios.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Z-Score
In the realm of statistics, a Z-score is a way to measure how many standard deviations a data point is from the mean of a dataset. Think of it as a way to standardize data that lets you see how different or extreme a particular value is compared to the average.
  • The Z-score is calculated using the formula: \( Z = \frac{x-\mu}{\sigma} \).
  • Here, \(x\) is the value of interest, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.
  • A Z-score of 0 indicates that the data point is exactly average.
  • A positive Z-score means the data point is above the mean, while a negative one means it's below the mean.
Calculating the Z-score helps you understand the relative standing of a data point. By knowing the Z-score, you can determine the probability or percentage of data points that fall below or above this value within a normal distribution.
Role of Standard Deviation
Standard deviation, depicted as \(\sigma\), is the measure of data dispersion in a dataset. It tells you how much individual data points differ from the mean.
  • If the standard deviation is small, it means the data points are close to the mean.
  • If it鈥檚 large, the data points are spread out over a wider range of values.
In the context of a Z-score, the standard deviation serves as the unit for measuring how far a particular value is from the mean.

For example:
  • In a dataset with mean \(\mu = 55\) and standard deviation \(\sigma = 7\), knowing \(\sigma\) helps calculate how much a specific value like \(x = 58\) deviates from the mean in terms of standard deviations.
This concept is pivotal to understanding normal distributions since the shape of the curve is determined by how spread out the data is.
Using the Z-Table
A Z-table, often called the standard normal distribution table, is crucial for finding probabilities related to Z-scores. It shows the percentage or probability of values being less than a specific Z-score in a standard normal distribution.
  • The Z-table typically gives the area or probability to the left of a Z-score.
  • To find the probability of a value being 'to the right' of a Z-score, you subtract the table value from 1.
  • For instance, if the Z-score is 1.96, and the Z-table provides 0.975, the probability of scores greater than 1.96 is \(1 - 0.975 = 0.025\).
Utilizing the Z-table allows for precise calculation of where a certain data point falls within the distribution, making it indispensable in statistics for tasks such as hypothesis testing and confidence interval estimation.

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Most popular questions from this chapter

The Bank of Connecticut issues Visa and MasterCard credit cards. It is estimated that the balances on all Visa credit cards issued by the Bank of Connecticut have a mean of \(\$ 845\) and a standard deviation of \(\$ 270\). Assume that the balances on all these Visa cards follow a normal distribution. a. What is the probability that a randomly selected Visa card issued by this bank has a balance between \(\$ 1000\) and \(\$ 1440 ?\) h. What percentage of the Visa cards issued by this bank have a balance of \(\$ 730\) or more?

Suppose you are conducting a binomial experiment that has 15 trials and the probability of success of \(.02\). According to the sample size requirements, you cannot use the normal distribution to approximate the binomial distribution in this situation. Use the mean and standard deviation of this binomial distribution and the empirical rule to explain why there is a problem in this situation. (Note: Drawing the graph and marking the values that correspond to the empirical rule is a good way to start.)

Tommy Wait, a minor league baseball pitcher, is notorious for taking an excessive amount of time between pitches. In fact, his times between pitches are normally distributed with a mean of 36 seconds and a standard deviation of \(2.5\) seconds. What percentage of his times between pitches are a. longer than 39 seconds? b. between 29 and 34 seconds?

Find the value of \(z\) so that the area under the standard normal curve a. from 0 to \(z\) is (approximately) \(.1965\) and \(z\) is positive b. between 0 and \(z\) is (approximately) 2740 and \(z\) is negative c. in the left tail is (approximately) \(.2050\) d. in the right tail is (approximately). 1053

A machine at Kasem Steel Corporation makes iron rods that are supposed to be 50 inches long. However, the machine does not make all rods of exactly the same length. It is known that the probability distribution of the lengths of rods made on this machine is normal with a mean of 50 inches and a standard deviation of \(.06\) inch. The rods that are either shorter than \(49.85\) inches or longer than \(50.15\) inches are discarded. What percentage of the rods made on this machine are discarded?
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