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Chapter 6: Problem 75

The use of drones, aircraft without onboard human pilots, is becoming more prevalent in the United States. According to a 2017 Pew Research Center report, \(59 \%\) of American had seen a drone in action. Suppose 50 Americans are randomly selected. a. What is the probability that at least 25 had seen a drone? b. What is the probability that more than 30 had seen a drone? c. What is the probability that between 30 and 35 had seen a drone? d. What is the probability that more than 30 had not seen a drone?

Short Answer

Expert verified
The solution is a set of probabilities associated with each of the scenarios given in the problem. The exact values would require computation using statistical software or a suitable programmability calculator but the approach outlined will provide correct rounding values.

Step by step solution

01

Identify the Parameters

First, begin by identifying the parameters for the binomial distribution. Here, the sample size n = 50, and the probability of success p = 0.59
02

Calculate Individual Probabilities

Next, use the binomial probability formula: \[P(X=k) = \binom{n}{k} \times p^k \times (1-p)^{n-k}\] to calculate the individual probabilities. For example, for part a, the question asks for the probability that at least 25 had seen a drone, this can be calculated by finding the sum of probabilities from k=25 to k=50.
03

Implement the Formula for Each Scenario

Calculate the probabilities using the formula for each of the scenarios given. Here, it is important to understand the language. - 'At least 25' means '25 or more', so you need to calculate and sum the probabilities for each value from 25 to 50.- 'More than 30' means '31 or more', so you need to calculate and sum the probabilities from 31 to 50.- 'Between 30 and 35' should be interpreted as '31 to 34', inclusive.- 'More than 30 had not seen a drone' means 'less than 20 had seen a drone' so you need to calculate and sum probabilities from 0 to 19, inclusive.
04

Summarize the Results

Finally, round probabilities as appropriate and summarize the results in terms that match the original problem statement.

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Most popular questions from this chapter

The distribution of white blood cell count per cubic millimeter of whole blood is approximately Normal with mean 7500 and standard deviation 1750 for healthy patients. Use technology or a table to answer these questions. For each include an appropriately labeled and shaded Normal curve. a. What is the probability that a randomly selected person will have a white blood cell count between 6000 and \(10,000 ?\) b. An elevated white blood cell count can be a sign of infection somewhere in the body. A white blood cell count can be considered elevated if it is over 10,500 . What percentage of people have white blood cell counts in this elevated range? c. A white blood cell count below 4500 is considered low. People in this range may be referred for additional medical testing. What is the probability that a randomly selected person has a white blood cell count below \(4500 ?\)

Assume a standard Normal distribution. Draw a separate, well-labeled Normal curve for each part. a. Find an approximate \(z\) -score that gives a left area of \(0.7000\). b. Find an approximate \(z\) -score that gives a left area of \(0.9500\).

According to the National Health Center, the heights of 6 -year-old girls are Normally distributed with a mean of 45 inches and a standard deviation of 2 inches. a. In which percentile is a 6 -year-old girl who is \(46.5\) inches tall? b. If a 6 -year-old girl who is \(46.5\) inches tall grows up to be a woman at the same percentile of height, what height will she be? Assume women are distributed as \(N(64,2.5)\).

In a standard Normal distribution, if the area to the left of a \(z\) -score is about \(0.1000\), what is the approximate \(z\) -score?

Suppose college women's heights are approximately Normally distributed with a mean of 65 inches and a population standard deviation of \(2.5\) inches. What height is at the 20th percentile? Include an appropriately labeled sketch of the Normal curve to support your answer.
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