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Chapter 4: Problem 55

Suppose only \(75 \%\) of all drivers in a certain state regularly wear a seat belt. A random sample of 500 drivers is selected. What is the probability that a. Between 360 and 400 (inclusive) of the drivers in the sample regularly wear a seat belt? b. Fewer than 400 of thore in the sample regularly wear a seat belt?

Short Answer

Expert verified
a. Probability is 0.9345. b. Probability is 0.9951.

Step by step solution

01

Identify the Distribution

The scenario describes a binomial setting because each driver either wears a seat belt or does not. We consider 500 drivers, with each having a probability of wearing a seat belt at 0.75.
02

Check for Normal Approximation

With a large sample size (n=500), we use normal approximation to the binomial. The conditions np and nq are both greater than 5: \( np = 500 \times 0.75 = 375 \) and \( nq = 500 \times 0.25 = 125 \). Thus, normal approximation is appropriate.
03

Calculate Mean and Standard Deviation

For a binomial distribution, the mean \( \mu \) is given by \( np \) and the standard deviation \( \sigma \) by \( \sqrt{npq} \). So, \( \mu = 375 \) and \( \sigma = \sqrt{500 \times 0.75 \times 0.25} \approx 9.68 \).
04

Convert to Standard Normal Variable for Part (a)

We calculate the z-scores for 360 and 400: \[ z = \frac{360 - 375}{9.68} \approx -1.55 \] and \[ z = \frac{400 - 375}{9.68} \approx 2.58 \].
05

Find Probability for Part (a)

Using standard normal distribution tables or a calculator, find the probabilities corresponding to the z-scores. For \( z = -1.55 \), \( P(Z < -1.55) \approx 0.0606 \), and for \( z = 2.58 \), \( P(Z < 2.58) \approx 0.9951 \). Thus, \( P(360 \leq X \leq 400) \approx 0.9951 - 0.0606 = 0.9345 \).
06

Convert to Standard Normal Variable for Part (b)

For fewer than 400, simply use the earlier z-score for 400: \[ z = \frac{400 - 375}{9.68} \approx 2.58 \].
07

Find Probability for Part (b)

Using the standard normal distribution table, find \( P(X < 400) = P(Z < 2.58) \approx 0.9951 \).

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Key Concepts

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

Binomial Distribution
When we talk about the binomial distribution, we're dealing with scenarios where there are fixed numbers of trials, and each trial has only two possible outcomes, like success or failure. For example, in the exercise, each driver can either wear a seat belt (success) or not (failure). We conduct our trials with 500 drivers, which is our fixed number of trials.

A binomial distribution is defined by two parameters:
  • \( n \): the number of trials (500 drivers in this case).
  • \( p \): the probability of success on each trial (0.75 probability of wearing a seat belt).
Using these parameters, this distribution helps us model and predict the number of successes over many trials. It provides a coherent way to calculate probabilities like the likelihood of a certain number of drivers wearing their seat belts.
Normal Approximation
In statistics, sometimes it's easier to work with the normal distribution rather than the binomial distribution, especially when sample sizes are large. This is where normal approximation comes into play. It simplifies calculations by allowing us to use the normal distribution to estimate binomial probabilities.

However, there are certain conditions for this approximation to be valid:
  • \( np \): expected number of successes, must be greater than 5.
  • \( nq \): expected number of failures, must also be greater than 5.
In our exercise, these conditions are met since \( np = 375 \) and \( nq = 125 \). Thus, using a normal approximation allows us to continue with simpler calculations while maintaining accurate probability estimations.
Standard Normal Distribution
The standard normal distribution is a special case of the normal distribution. It comes with a mean of 0 and a standard deviation of 1. This distribution is useful because it provides a basis for comparison and helps in finding probabilities for any normal distribution by standardizing the values.

To transform our data points to fit the standard normal distribution, we calculate z-scores. This transformation allows us to consult standardized distribution tables and find the probabilities of being within a certain range of values, as we see in our exercise. Before doing this, we first identify the binomial mean and standard deviation, which help in converting actual values into equivalent z-scores.
Z-score Calculation
Z-score calculation is a method to determine how far and in what direction a data point is from the mean, measured in units of standard deviation. The formula for calculating a z-score is:\[ z = \frac{X - \mu}{\sigma}\]where:
  • \( X \) is the value you're analyzing (like the number of drivers).
  • \( \mu \) is the mean of the sample.
  • \( \sigma \) is the standard deviation.
In our example, calculating the z-scores for values 360 and 400 helps determine the probability that a certain number of drivers wear seat belts. By finding the z-score, we can then use a standard normal distribution table to find the associated probability. This method bridges the gap between raw data and the probabilistic insights drawn from it.

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Most popular questions from this chapter

Consider babies born in the "normal" rangge of \(37-43\) weeks gestational age. Extensive data supports the assumption that for such babies born in the Lnited States, birth weight is normally distributed with mean \(3432 \mathrm{~g}\) and standard deviation \(482 \mathrm{~g}\). [The article "Are Babies Normal?" (The American Statistician (1999): 298-302) analyzed data from a particu" lar year; for a sensible choice of class intervals, a histogram did not look at all normal but after further investigations it was detenmined that this was due to some bospitals measuring weight in grams and others measuring to the nearest ounce and then converting to grams. A modified choice of class intervals that allowed for this gave a histogram that was well described by a normal distribution.] a. What is the probability that the birth weight of a randomly selected baby of this type exceeds 4000 grams? Is between 3000 and 4000 grams? b. What is the probability that the birth weight of a randomly selected baby of this type is either less than 2000 grams or greater than 5000 grams? c. What is the probability that the birth weight of a randoenly selected baby of this type exceeds \(7 \mathrm{lb}\) ? d. How would you characterize the most extreme .1\% of all birth weights? e. If \(X\) is a random variahle with a normal distribution and \(a\) is a numerical constant \((a \neq 0\) ), then \(Y-a X\) also has a normal distribution. Use this to determine the distribution of birth weight expressed in pounds (shape, mean, and standard deviation), and then recalculate the probability

Spray drift is a constant concern for pesticide applicators and agricultural producers. The inverse relationship between droplet size and drift potential is well known. The paper "Effects of \(2,4-\mathrm{D}\) Formulation and Quinclorac on Spray Droplet Size and Deposition" (Weed Technology, 2005: \(1030-1036\) ) investigated the effects of herbicide formulation on spray atomization. A figure in the paper suggested the normal distribution with mean \(1050 \mu \mathrm{m}\) and standard deviation \(150 \mu \mathrm{m}\) was a reasonable model for droplet size for water (the "control treatment") sprayed through a \(760 \mathrm{ml} / \mathrm{min}\) nozzle. a. What is the probability that the size of a single droplet is less than \(1500 \mu \mathrm{m}\) ? At least \(1000 \mu \mathrm{m}\) ? b. What is the probability that the size of a single droplet is between 1000 and \(1500 \mu \mathrm{m}\) ? c. How would you characterize the smallest \(2 \%\) of all droplets? d. If the sizes of five independently selected droplets are measured, what is the probability that at least one exceeds \(1500 \mu \mathrm{m}\) ?

a. If a normal distribution has \(\mu=30\) and \(\sigma=5\), what is the 91 st percentile of the distribution? b. What is the 6th percentile of the distribution? c. The width of a line etched on an integrated circuit chip is normally distributed with mean \(3.000 \mu \mathrm{m}\) and standard deviation \(.140\). What width value separates the widest \(10 \%\) of all such lines from the other \(90 \%\) ?

Let the ordered sample observations be denoted by \(y_{1}\), \(y_{2}, \ldots, y_{n}\) ( \(y_{1}\) being the smallest and \(y_{n}\) the largest). Our suggested check for normality is to plot the \(\left(\Phi^{-1}((i-.5) / n), y_{i}\right)\) pairs. Suppose we believe that the observations come from a distribution with mean 0 , and let \(w_{1}, \ldots, w_{n}\) be the ordered absolute values of the \(x_{i}\) s. A half-normal plot is a probability plot of the \(w_{i} s\). More specifically, since \(P(|Z| \leq w)=\) \(P(-w \leq Z \leq w)=2 \Phi(w)-1\), a half-normal plot is a plot of the \(\left(\Phi^{-1}\\{[(i-5) / n+1] / 2\\}, w_{i}\right)\) pairs. The virtue of this plot is that small or large outliers in the original sample will now appear only at the upper end of the plot rather than at both ends. Construct a half-normal plot for the following sample of measurement errors, and comment: \(-3.78,-1.27,1.44\), \(-39,12.38,-43.40,1.15,-3.96,-2.34,30.84\).

Let \(X\) denote the voltage at the output of a microphone, and suppose that \(X\) has a uniform distribution on the interval from \(-1\) to 1 . The voltage is processed by a "hard limiter" with cutoff values \(-.5\) and \(.5\), so the limiter output is a random variable \(Y\) related to \(X\) by \(Y=X\) if \(|X| \leq .5, Y=.5\) if \(X>.5\), and \(Y=-.5\) if \(X<-.5\). a. What is \(P(Y=.5)\) ? b. Obtain the cumulative distribution function of \(Y\) and graph it.
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