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Chapter 6: Problem 1

In testing the assumption that the probability of a baby boy is \(0.512\), a geneticist obtains a random sample of 1000 births and finds that 502 of them are boys. Using the continuity correction, describe the area under the graph of a normal distribution corresponding to the following. (For example, the area corresponding to "the probability of at least 502 boys" is this: the area to the right of \(501.5 .\) ) a. The probability of 502 or fewer boys b. The probability of exactly 502 boys c. The probability of more than 502 boys

Short Answer

Expert verified
a. 0.2743, b. 0.0196, c. 0.7257

Step by step solution

01

Understand the Problem

We need to determine the areas under the normal distribution curve for different scenarios, using the continuity correction.
02

Define the Mean and Standard Deviation

The mean of a binomial distribution is given by \(\text{np}\) and the standard deviation is \(\text{sqrt(np(1-p))}\). Here, \(n = 1000\) and \(p = 0.512\). Thus, \[ \text{mean} (\text{np}) = 1000 \times 0.512 = 512 \] and \[ \text{standard deviation} (\text{sqrt(np(1-p))}) = \text{sqrt}(1000 \times 0.512 \times 0.488) \approx 15.81 \]
03

Probability of 502 or Fewer Boys

To find the probability of 502 or fewer boys, use the continuity correction to adjust the apparent range to the left of 502. Thus, the equivalent z-score calculation is for \(P(X \leq 502.5)\). Calculate the z-score: \[ z = \frac{502.5 - 512}{15.81} \approx -0.60 \] Use standard normal distribution tables or a calculator to find \(P(Z \leq -0.60)\). This corresponds to an area of approximately 0.2743.
04

Probability of Exactly 502 Boys

To find the probability of exactly 502 boys, use the continuity correction to consider the interval from 501.5 to 502.5. Thus, we calculate \( P(501.5 < X \leq 502.5) \). Calculate the z-scores: \[ z_1 = \frac{501.5 - 512}{15.81} \approx -0.66, \] \[ z_2 = \frac{502.5 - 512}{15.81} \approx -0.60 \] Find the area between these z-scores in the standard normal distribution table. The corresponding area is approximately \(P(-0.66 \leq Z \leq -0.60) \approx 0.0196 \).
05

Probability of More Than 502 Boys

To find the probability of more than 502 boys, again employ the continuity correction. So calculate the z-score for \(P(X > 502.5)\). Calculate the z-score: \[ z = \frac{502.5 - 512}{15.81} \approx -0.60 \] Use standard normal distribution tables or a calculator to find \(P(Z > -0.60)\) which is equivalent to \(1 - P(Z \leq -0.60) \approx 1 - 0.2743 = 0.7257.\)

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Key Concepts

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

Probability Distribution
In statistics, a probability distribution describes how the values of a random variable are distributed. It provides the probabilities of different possible outcomes. There are various types of probability distributions, including the normal distribution and the binomial distribution.

The normal distribution, also known as the Gaussian distribution, is a continuous probability distribution. It is symmetrical and characterized by its bell-shaped curve. The mean and standard deviation of the data determine its shape and spread.

The binomial distribution, in contrast, is a discrete probability distribution. It describes the number of successes in a fixed number of independent trials, each with the same probability of success. For example, in the context of the exercise, the probability of having a baby boy (success) in each birth (trial) can be modeled using a binomial distribution.
Continuity Correction
When working with discrete data in a context that assumes a continuous distribution, a continuity correction is applied to better approximate the probability. This is particularly useful when transforming a binomial distribution (which is discrete) into a normal distribution (which is continuous).

In our exercise, the continuity correction adjusts the calculations by 0.5 units, converting discrete values (like exactly 502 boys) into an interval (like 501.5 to 502.5). This helps improve the approximation made by the normal distribution for the binomial probabilities. For instance, to calculate the probability of having 502 or fewer boys, we use 502.5 instead of 502.
Z-score Calculation
A z-score measures how many standard deviations an element is from the mean. It is a fundamental concept in statistics for understanding where a value lies within a distribution. The z-score is given by the formula: \[ z = \frac{X - \text{mean}}{\text{standard deviation}} \]

In the exercise, we need to calculate the z-scores to determine areas under the normal distribution curve. For example, to find the z-score for 502.5 boys, given a mean of 512 and a standard deviation of approximately 15.81, we plug these values into our formula. This yields: \[ z = \frac{502.5 - 512}{15.81} \approx -0.60 \] Using the z-score, we then refer to standard normal distribution tables or a calculator to find the corresponding probability.
Binomial Distribution
The binomial distribution is a model for the number of successes in a fixed number of independent and identically distributed Bernoulli trials. Each trial has two possible outcomes: success or failure. The binomial distribution is defined by two parameters: the number of trials (n) and the probability of success on a given trial (p).

In our exercise, each birth is a trial, and the probability of a boy (success) is 0.512. With 1000 total births, we can calculate the mean and standard deviation of the binomial distribution as follows: \[ \text{mean} = n \times p = 1000 \times 0.512 = 512 \] \[ \text{standard deviation} = \text{sqrt}(n \times p \times (1 - p)) = \text{sqrt}(1000 \times 0.512 \times 0.488) \approx 15.81 \]

These parameters allow us to transform the discrete binomial probabilities into a continuous normal approximation, using a continuity correction and z-scores.

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Most popular questions from this chapter

Use the data in the table below for sitting adult males and females (based on anthropometric survey data from Gordon, Churchill, et al.). These data are used often in the design of different seats, including aircraft seats, train seats, theater seats, and classroom seats. (Hint: Draw a graph in each case.) $$ \begin{aligned} &\text { Sitting Back-to-Knee Length (inches) }\\\ &\begin{array}{l|c|c|c} \hline & \text { Mean } & \text { St. Dev. } & \text { Distribution } \\ \hline \text { Males } & 23.5 \text { in. } & 1.1 \text { in. } & \text { Normal } \\ \hline \text { Females } & 22.7 \text { in. } & 1.0 \text { in. } & \text { Normal } \\ \hline \end{array} \end{aligned} $$ Instead of using \(0.05\) for identifying significant values, use the criteria th?t a value \(x\) is significantly high if \(P(x\) or greater \() \leq 0.01\) and a value is significantly low if \(P(x\) or less \() \leq 0.01\). Find the back-to- knee lengths for males, separating significant values from those that are not significant. Using these criteria, is a male back-to-knee length of 26 in. significantly high?

When a water taxi sank in Baltimore's Inner Harbor, an investigation revealed that the safe passenger load for the water taxi was \(3500 \mathrm{lb}\). It was also noted that the mean weight of a passenger was assumed to be \(140 \mathrm{lb}\). Assume a "worst-case" scenario in which all of the passengers are adult men. Assume that weights of men are normally distributed with a mean of \(188.6 \mathrm{lb}\) and a standard deviation of \(38.9 \mathrm{lb}\) (based on Data Set 1 "Body Data" in Appendix B). a. If one man is randomly selected, find the probability that he weighs less than \(174 \mathrm{lb}\) (the new value suggested by the National Transportation and Safety Board). b. With a load limit of \(3500 \mathrm{lb}\), how many male passengers are allowed if we assume a mean weight of \(140 \mathrm{lb}\) ? c. With a load limit of \(3500 \mathrm{lb}\), how many male passengers are allowed if we assume the updated mean weight of \(188.6 \mathrm{lb}\) ? d. Why is it necessary to periodically review and revise the number of passengers that are allowed to board?

Common tests such as the SAT, ACT, Law School Admission test (LSAT), and Medical College Admission Test (MCAT) use multiple choice test questions, each with possible answers of \(a, b, c, d, e\), and each question has only one correct answer. We want to find the probability of getting at least 25 correct answers for someone who makes random guesses for answers to a block of 100 questions. If we plan to use the methods of this section with a normal distribution used to approximate a binomial distribution, are the necessary requirements satisfied? Explain.

Before every flight, the pilot must verify that the total weight of the load is less than the maximum allowable load for the aircraft. The Bombardier Dash 8 aircraft can carry 37 passengers, and a flight has fuel and baggage that allows for a total passenger load of \(6200 \mathrm{lb}\). The pilot sees that the plane is full and all passengers are men. The aircraft will be overloaded if the mean weight of the passengers is greater than \(6200 \mathrm{lb} / 37=167.6 \mathrm{lb}\). What is the probability that the aircraft is overloaded? Should the pilot take any action to correct for an overloaded aircraft? Assume that weights of men are normally distributed with a mean of \(189 \mathrm{lb}\) and a standard deviation of \(39 \mathrm{lb}\) (based on Data Set 1 "Body Data" in Appendix B).

Use the data in the table below for sitting adult males and females (based on anthropometric survey data from Gordon, Churchill, et al.). These data are used often in the design of different seats, including aircraft seats, train seats, theater seats, and classroom seats. (Hint: Draw a graph in each case.) $$ \begin{aligned} &\text { Sitting Back-to-Knee Length (inches) }\\\ &\begin{array}{l|c|c|c} \hline & \text { Mean } & \text { St. Dev. } & \text { Distribution } \\ \hline \text { Males } & 23.5 \text { in. } & 1.1 \text { in. } & \text { Normal } \\ \hline \text { Females } & 22.7 \text { in. } & 1.0 \text { in. } & \text { Normal } \\ \hline \end{array} \end{aligned} $$ For females, find the first quartile \(Q_{1}\), which is the length separating the bottom \(25 \%\) from the top \(75 \%\).
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