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Ch. 5 - Normal Probability Distributions
Larson - Elementary Statistics: Picturing the World 8th Edition
Larson8th EditionElementary Statistics: Picturing the WorldISBN: 9780137493470Not the one you use?Change textbook
Chapter 5, Problem 5.T.3c

In Exercises 5 and 6, determine whether you can use a normal distribution to approximate the binomial distribution. If you can, use the normal distribution to approximate the indicated probabilities and sketch their graphs. If you cannot, explain why and use a binomial distribution to find the indicated probabilities.


A survey of U.S. undergraduates found that 37% of those attending in-state colleges would prefer to take a job in a different state after graduation. You randomly select 18 U.S. undergraduates attending in-state colleges. Find the probability that the number who would prefer to take a job in a different state after graduation is (c) at least 10. Identify any unusual events. Explain.

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Step 1: Determine if the normal distribution can be used to approximate the binomial distribution. For this, check the conditions: (1) The sample size (n) must be large enough, and (2) both np and n(1-p) must be greater than or equal to 5. Here, n = 18 and p = 0.37. Calculate np = 18 × 0.37 and n(1-p) = 18 × (1 - 0.37).
Step 2: If the conditions are satisfied, proceed to approximate the binomial distribution using the normal distribution. The mean (μ) and standard deviation (σ) of the binomial distribution are given by μ = np and σ = √(np(1-p)). Calculate these values using the given n and p.
Step 3: To find the probability that at least 10 students prefer to take a job in a different state, use the continuity correction. Convert the discrete value 'at least 10' to a continuous value by considering P(X ≥ 10) as P(X > 9.5) in the normal distribution.
Step 4: Standardize the value 9.5 using the z-score formula: z = (X - μ) / σ, where X is the value of interest, μ is the mean, and σ is the standard deviation. Calculate the z-score for X = 9.5.
Step 5: Use the standard normal distribution table (or a calculator) to find the probability corresponding to the calculated z-score. Subtract this probability from 1 to find the probability of P(X ≥ 10). Sketch the graph of the normal distribution curve, marking the area corresponding to P(X ≥ 10). Identify any unusual events by comparing the probability to a threshold (e.g., 0.05 for unusual events).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Binomial Distribution

The binomial distribution models the number of successes in a fixed number of independent Bernoulli trials, each with the same probability of success. In this context, it applies to the scenario of selecting undergraduates who prefer to take a job in a different state, where each selection can be viewed as a trial with two outcomes: preferring a job out of state or not.
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Mean & Standard Deviation of Binomial Distribution

Normal Approximation to the Binomial

The normal approximation to the binomial distribution is applicable when the number of trials is large, and both the probability of success and failure are not too close to 0 or 1. Specifically, the rule of thumb is that both np and n(1-p) should be greater than 5. This allows us to use the normal distribution to estimate probabilities for binomial scenarios, simplifying calculations.
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Probability Calculation

Calculating probabilities involves determining the likelihood of a specific outcome occurring within a given distribution. For the binomial distribution, this can be done using the binomial probability formula, while for the normal approximation, the z-score formula is used to find probabilities under the normal curve. Understanding how to compute these probabilities is essential for answering the question regarding the preferences of undergraduates.
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Related Practice
Textbook Question

Assume the machine shifts and is filling the vials with a mean amount of 9.96 milligrams and a standard deviation of 0.05 milligram. You select five vials and find the mean amount of compound added.


c. Which is more sensitive to a shift of parameters—an individual random selection or a randomly selected sample mean?

Textbook Question

Use technology to find the standard deviation of the set of 36 sample means. How does it compare with the standard deviation of the ages found in Exercise 5? Does this agree with the result predicted by the Central Limit Theorem?

Textbook Question

Assume the machine shifts and the distribution of the amount of the compound added now has a mean of 9.96 milligrams and a standard deviation of 0.05 milligram. You select one vial and determine how much of the compound was added.



a. What is the probability that you select a vial that is within the acceptable range (in other words, you do not detect that the machine has shifted)? (See figure.)

Textbook Question

In Exercises 5 and 6, determine whether you can use a normal distribution to approximate the binomial distribution. If you can, use the normal distribution to approximate the indicated probabilities and sketch their graphs. If you cannot, explain why and use a binomial distribution to find the indicated probabilities.


A survey of U.S. undergraduates found that 37% of those attending in-state colleges would prefer to take a job in a different state after graduation. You randomly select 18 U.S. undergraduates attending in-state colleges. Find the probability that the number who would prefer to take a job in a different state after graduation is (b) less than 5. Identify any unusual events. Explain.

Textbook Question

Assume the machine shifts and is filling the vials with a mean amount of 9.96 milligrams and a standard deviation of 0.05 milligram. You select five vials and find the mean amount of compound added.


a. What is the probability that you select a sample of five vials that has a mean that is within the acceptable range? (See figure.)

Textbook Question

In Exercises 2–4, the random variable x is normally distributed with mean mu= 18 and standard deviation sigma 7.6


Find the value of x that has 88.3% of the distribution’s area to its left.