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Chapter 6: Problem 19

Find the \(z\) value for each of the following \(x\) values for a normal distribution with \(\mu=30\) and \(\sigma=5\). a. \(x=39\) b. \(x=19\) c. \(x=24\) d. \(x=44\)

Short Answer

Expert verified
The z-scores for the given \(x\) values are: a. \(Z_{39}=1.8\), b. \(Z_{19}=-2.2\), c. \(Z_{24}=-1.2\), d. \(Z_{44}=2.8\)

Step by step solution

01

Understand z-score and its formula

The z-score is calculated using the formula: \(Z = (X - \mu) / \sigma\). Where \(Z\) is the z-score, \(X\) is the value from the dataset, \(\mu\) is the mean, and \(\sigma\) is the standard deviation. In this case, the mean :\(\mu\) is 30 and the standard deviation :\(\sigma\) is 5.
02

Calculate z-score for x=39

Substitute \(X = 39\) into the z-score formula. The z-score is: \(Z_{39} = (39 - 30) / 5 = 1.8\)
03

Calculate z-score for x=19

Substitute \(X = 19\) into the z-score formula. The z-score is: \(Z_{19} = (19 - 30) / 5 = -2.2\)
04

Calculate z-score for x=24

Substitute \(X = 24\) into the z-score formula. The z-score is: \(Z_{24} = (24 - 30) / 5 = -1.2\)
05

Calculate z-score for x=44

Substitute \(X = 44\) into the z-score formula. The z-score is: \(Z_{44} = (44 - 30) / 5 = 2.8\)

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Key Concepts

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

Normal Distribution
The normal distribution is one of the most important probability distributions in statistics. It's bell-shaped and symmetric, meaning it looks the same on the left and right of the central peak. This distribution is defined by two parameters: the mean (\( \mu \) ) and the standard deviation (\( \sigma \) ). The mean is the center of the distribution, where most data points are clustered, while the standard deviation indicates how spread out the data is.

Some key features of the normal distribution include:
  • It is unimodal, having a single peak.
  • 68% of the data approximately falls within one standard deviation from the mean.
  • 95% of the data falls within two standard deviations from the mean.
  • Almost all (99.7%) data points fall within three standard deviations.
Understanding the shape and characteristics of the normal distribution helps in analyzing data and predicting probabilities.
Mean and Standard Deviation
The mean and standard deviation are fundamental to understanding data distribution, especially in the context of a normal distribution. They are vital for calculating the z-score.

The mean, denoted as \( \mu \), is the average of all the data points. It is calculated by summing all values and dividing by the count. In our example, the mean is given as 30.
  • It gives us a central point around which the other data points are distributed.
  • In a normal distribution, this is the peak of the bell curve.


The standard deviation, denoted as \( \sigma \), measures the spread or dispersion of a set of data points. For this exercise, it is given as 5.
  • A smaller \( \sigma \) means data is closely clustered around the mean.
  • A larger \( \sigma \) indicates data is spread out over a wider range.
Together, the mean and standard deviation provide a complete view of the data distribution's center and spread.
Z-Score Formula
The z-score formula is a crucial tool used to determine how far a specific data point is from the mean, in terms of standard deviations. The formula is expressed as:

\[ Z = \frac{X - \mu}{\sigma} \] where:
  • \( Z \) is the z-score.
  • \( X \) is the data point.
  • \( \mu \) is the mean.
  • \( \sigma \) is the standard deviation.
The z-score tells us whether a data point is typical for the dataset or unusual.

  • A z-score of 0 means the data point is exactly at the mean.
  • Positive z-scores indicate points above the mean.
  • Negative z-scores indicate points below the mean.
Using the z-score is helpful when comparing values from different datasets that may have different means and standard deviations.
Step-by-Step Solution
To find the z-scores for a given set of data points, follow a clear, step-by-step process.

Step 1: Identify the mean \( \mu \) and standard deviation \( \sigma \). In this example, \( \mu = 30 \) and \( \sigma = 5 \).

Step 2: Pick the first data point, \( X \). For example, suppose \( X = 39 \). Use the z-score formula:
\[ Z = \frac{39 - 30}{5} = 1.8 \]. This tells us 39 is 1.8 standard deviations above the mean.

Step 3: Repeat this calculation for each data point:
  • For \( X = 19 \), \[ Z = \frac{19 - 30}{5} = -2.2 \]. This indicates it's 2.2 standard deviations below the mean.
  • For \( X = 24 \), \[ Z = \frac{24 - 30}{5} = -1.2 \]. This shows it's 1.2 standard deviations below the mean.
  • For \( X = 44 \), \[ Z = \frac{44 - 30}{5} = 2.8 \]. This shows it's 2.8 standard deviations above the mean.
Each calculated z-score offers insight into the relative position of the data point within the distribution.

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Most popular questions from this chapter

At Jen and Perry Ice Cream Company, the machine that fills 1-pound cartons of Top Flavor ice cream is set to dispense 16 ounces of ice cream into every carton. However, some cartons contain slightly less than and some contain slightly more than 16 ounces of ice cream. The amounts of ice cream in all such cartons have a normal distribution with a mean of 16 ounces and a standard deviation of \(.18\) ounce. a. Find the probability that a randomly selected carton contains \(16.20\) to \(16.50\) ounces of ice cream. b. What percentage of such cartons contain less than \(15.70\) ounces of ice cream? c. Is it possible for a carton to contain less than \(15.20\) ounces of ice cream? Explain.

Let \(x\) denote the time taken to run a road race. Suppose \(x\) is approximately normally distributed with a mean of 190 minutes and a standard deviation of 21 minutes. If one runner is selected at random, what is the probability that this runner will complete this road race a. in less than 160 minutes? b. in 215 to 245 minutes?

One of the cars sold by Walt's car dealership is a very popular subcompact car called the 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?

Fast Auto Service guarantees that the maximum waiting time for its customers is 20 minutes for oil and lube service on their cars. It also guarantees that any customer who has to wait longer than 20 minutes for this service will receive a \(50 \%\) discount on the charges. It is estimated that the mean time taken for oil and lube service at this garage is 15 minutes per car and the standard deviation is \(2.4\) minutes. Suppose the time taken for oil and lube service on a car follows a normal distribution. a. What percentage of customers will receive a \(50 \%\) discount on their charges? b. Is it possible that it may take longer than 25 minutes for oil and lube service? Explain.

At Jen and Perry Ice Cream Company, a machine fills 1 -pound cartons of Top Flavor ice cream. The machine can be set to dispense, on average, any amount of ice cream into these cartons. However, the machine does not put exactly the same amount of ice cream into each carton; it varies from carton to carton. It is known that the amount of ice cream put into each such carton has a normal distribution with a standard deviation of \(.18\) ounce. The quality control inspector wants to set the machine such that at least \(90 \%\) of the cartons have more than 16 ounces of ice cream. What should be the mean amount of ice cream put into these cartons by this machine?
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