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Chapter 6: Problem 7

Assume that \(x\) has a normal distribution with the specified mean and standard deviation. Find the indicated probabilities. $$ P(50 \leq x \leq 70) ; \mu=40 ; \sigma=15 $$

Short Answer

Expert verified
The probability is approximately 0.2286.

Step by step solution

01

Standardize the Variables

To find the probability for a normal distribution, we first need to convert our specified range of values for \(x\) into a standard normal distribution (Z-score). This standardization is done using the formula: \[ Z = \frac{x - \mu}{\sigma} \]where \( \mu = 40 \) and \( \sigma = 15 \).
02

Calculate Z-scores

First, compute the Z-score for \(x = 50\):\[ Z_1 = \frac{50 - 40}{15} = \frac{10}{15} = 0.67 \]Next, calculate the Z-score for \(x = 70\):\[ Z_2 = \frac{70 - 40}{15} = \frac{30}{15} = 2.00 \]
03

Use Z-table to Find Probabilities

Using the Z-table, find the probability corresponding to each Z-score. - For \(Z = 0.67\), the table gives a probability of approximately 0.7486.- For \(Z = 2.00\), the table gives a probability of approximately 0.9772.
04

Calculate the Probability for the Range

The probability that \(x\) is between 50 and 70 can be found by subtracting the probability of \(Z_1\) from the probability of \(Z_2\):\[ P(0.67 \leq Z \leq 2.00) = P(Z \leq 2.00) - P(Z \leq 0.67) = 0.9772 - 0.7486 = 0.2286 \]
05

Conclusion

Thus, the probability that a normally distributed random variable \(x\) is between 50 and 70, with a mean \(\mu = 40\) and a standard deviation \(\sigma = 15\), is 0.2286.

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Key Concepts

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

Mean and Standard Deviation
When dealing with a normal distribution, two key statistics are essential: the mean (bc) and the standard deviation (c3).
  • The **mean** is simply the average of all the values in the distribution. In your problem, this is 40, and it represents the central location of the data.
  • The **standard deviation** measures how spread out the values in the dataset are around the mean. A larger standard deviation means a wider spread of values. Here, the standard deviation is 15, indicating the data is spread over a moderate range around the mean.
These two parameters define the shape of the normal distribution curve. The mean locates the center, and the standard deviation determines the width. Understanding these helps predict probabilities and variations of data under the curve.
Z-score
The Z-score is a way of standardizing scores to compare them within the context of a normal distribution. Simply put, it measures how many standard deviations an element is from the mean.
  • To find the Z-score for a value, you use the formula: \[ Z = \frac{x - \mu}{\sigma} \]
  • This formula converts a normal distribution to a **standard normal distribution**, which has a mean of 0 and a standard deviation of 1.
Calculating a Z-score helps us understand the relative position of a value within the distribution. For example, a Z-score of 0 indicates the value is precisely at the mean of the distribution. If you have a Z-score of 2, the score is 2 standard deviations above the mean, hinting it's higher than most values in this normal distribution.
Probability
Probability in statistics is a measure of the likelihood that an event will occur. In the context of a normal distribution, we are often interested in the probability that a value falls within a certain range.
  • Using the Z-scores for the endpoints of your range, you interpret these scores with a **Z-table** to find corresponding probabilities.
  • The Z-table tells you the probability that a value is less than a given Z-score in the standard normal distribution.
Therefore, to find the probability that a variable falls between two Z-scores, you subtract the probability of the lower Z-score from the probability of the higher Z-score. This gives you the likelihood that a value falls within that range in the distribution.
Standard Normal Distribution
A standard normal distribution, often represented by a bell-shaped curve, has specific characteristics that make it valuable for statistical analysis. It has a mean of 0 and a standard deviation of 1. This simple structure helps statisticians make quick, relevant decisions about data.
  • When you standardize a normal distribution by converting raw scores into Z-scores, you effectively map the original data onto the standard normal distribution.
  • This makes it easier to find the probabilities associated with different ranges of data by using a Z-table.
The use of the standard normal distribution simplifies solving probability problems across various disciplines, since once a dataset is standardized, you don't need information about its original scale. You can handle probabilities universally.

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Most popular questions from this chapter

What percentage of the area under the normal curve lies (a) to the right of \(\mu ?\) (b) between \(\mu-2 \sigma\) and \(\mu+2 \sigma ?\) (c) to the right of \(\mu+3 \sigma ?\)

Find the \(z\) value described and sketch the area described. Find \(z\) such that \(95 \%\) of the standard normal curve lies to the right of \(z\).

Based on long experience, an airline found that about \(6 \%\) of the people making reservations on a flight from Miami to Denver do not show up for the flight. Suppose the airline overbooks this flight by selling 267 ticket reservations for an airplane with only 255 seats. (a) What is the probability that a person holding a reservation will show up for the flight? (b) Let \(n=267\) represent the number of ticket reservations. Let \(r\) represent the number of people with reservations who show up for the flight. Which expression represents the probability that a seat will be available for everyone who shows up holding a reservation? \(P(255 \leq r) ; \quad P(r \leq 255) ; \quad P(r \leq 267) ; \quad P(r=255)\) (c) Use the normal approximation to the binomial distribution and part (b) to answer the following question: What is the probability that a seat will be available for every person who shows up holding a reservation?

You need to compute the probability of 5 or fewer successes for a binomial experiment with 10 trials. The probability of success on a single trial is \(0.43 .\) Since this probability of success is not in the table, you decide to use the normal approximation to the binomial. Is this an appropriate strategy? Explain.

Consider a binomial experiment with 20 trials and probability \(0.45\) of success on a single trial. (a) Use the binomial distribution to find the probability of exactly 10 successes. (b) Use the normal distribution to approximate the probability of exactly 10 successes. (c) Compare the results of parts (a) and (b).
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