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Chapter 4: Problem 2

Suppose we have a random variable \(X\) where, for the values \(k\) \(=0, \ldots, 10,\) the associated probabilities are \(\left(\begin{array}{c}10 \\\ k\end{array}\right)(0.37)^{k}(0.63)^{10-k} .\) What is the mean of \(X ?\) (A) 0.37 (B) 0.63 (C) 3.7 (D) 6.3 (E) None of the above

Short Answer

Expert verified
The mean is 3.7, so the answer is (C).

Step by step solution

01

Identify the type of distribution

The given probabilities resemble the binomial distribution, represented by the formulai.e., \(P(X=k) = \binom{n}{k}p^k(1-p)^{n-k}\).Here, \(n = 10\), \(p = 0.37\), and thus, the variable \(X\) follows a binomial distribution.
02

Recall the mean formula for a binomial distribution

The mean \(\mu\) of a binomial distribution is given by the formula \(\mu = n \cdot p\).
03

Substitute the given values into the mean formula

Here, \(n = 10\) and \(p = 0.37\). Substitute these values into the mean formula:\[\mu = 10 \cdot 0.37 = 3.7\].
04

Compare with the given options

Compare the calculated mean \(3.7\) with the provided options: (A) 0.37 (B) 0.63 (C) 3.7 (D) 6.3 (E) None of the above.Clearly, the correct answer is (C) 3.7.

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Key Concepts

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

binomial distribution
A binomial distribution is a type of probability distribution. It describes the number of successes in a specific number of trials or experiments. Each trial is independent and can have only two outcomes: success or failure.
For example, if you flip a coin 10 times and count how many times you get heads, you are dealing with a binomial distribution. In this context, each flip is a trial, and getting heads is considered a success.
The formula for the binomial probability of getting exactly k successes in n trials is:
\[P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}\]
Here:
  • \(n\) is the number of trials
  • \(k\) is the number of successes
  • \(p\) is the probability of success on a single trial
  • \(1-p\) is the probability of failure
In the given exercise, you have 10 trials with a success probability of 0.37. The variable X represents the number of successes, making it a binomial random variable.
mean formula
The mean (or expected value) of a binomial distribution gives you the average number of successes you can expect in a set number of trials.
The formula for the mean \(\mu\) of a binomial distribution is:
\[\mu = n \cdot p\]
  • \(n\) is the number of trials
  • \(p\) is the probability of success on a single trial
In the exercise, you're given:\(n = 10\) and \(p = 0.37\). To find the mean, you substitute these values into the formula:
\[\mu = 10 \cdot 0.37 = 3.7\]
So, the average or mean number of successes you can expect after 10 trials is 3.7.
random variable
A random variable is a variable that can take on different values, each associated with a probability. In other words, it's a way to quantify randomness.
There are two types of random variables: discrete and continuous.
  • Discrete random variables: These can only take on a finite number of values. For example, the number of heads in 10 coin tosses is a discrete random variable.
  • Continuous random variables: These can take on an infinite number of values within a range. An example would be measuring the height of students in a class.
In our exercise, the random variable \(X\) represents the number of successes in 10 trials (or experiments). Since \(X\) can take on values like 0, 1, 2, and so on up to 10, it is a discrete random variable.
Knowing that \(X\) follows a binomial distribution helps us use the right formulas and methods to find quantities like the mean or the probability of specific outcomes.

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Most popular questions from this chapter

Mathematically speaking, casinos and life insurance companies make a profit because of (A) their understanding of sampling error and sources of bias. (B) their use of well-designed, well-conducted surveys and experiments. (C) their use of simulation of probability distributions. (D) the central limit theorem. (E) the law of large numbers.

There are two games involving flipping a fair coin. In the first game, you win a prize if you can throw between \(40 \%\) and \(60 \%\) heads. In the second game, you win if you can throw more than \(75 \%\) heads. For each game, would you rather flip the coin 50 times or 500 times? (A) 50 times for each game (B) 5 oo times for each game (C) 50 times for the first game, and 5 oo for the second (D) 5 oo times for the first game, and 50 for the second (E) The outcomes of the games do not depend on the number of flips.

An insurance company charges $$\$ 800$$ annually for car insurance. The policy specifies that the company will pay $$\$ 1000$$ for a minor accident and $$\$ 5000$$ for a major accident. If the probability of a motorist having a minor accident during the year is 0.2 and of having a major accident is 0.05 , how much can the insurance company expect to make on a policy? (A) $$\$ 200$$ (B) $$\$ 250$$ (C) $$\$ 300$$ (D) $$\$ 350$$ (E) $$\$ 450$$

Consider the following table of ages of U.S. senators: $$\begin{array}{|l|l|l|l|l|l|l|}\hline \text { Age (yr): } & <40 & 40-49 & 50-59 & 60-69 & 70-79 & >79 \\\\\hline \text { Number of senators: } & 5 & 30 & 36 & 22 & 5 & 2\\\\\hline\end{array}$$ What is the probability that a senator is under 70 years old given that he or she is at least 50 years old? (A) 0.580 (B) 0.624 (C) 0.643 (D) 0.892 (E) 0.969

In a 1974 "Dear Abby" letter, a woman lamented that she had just given birth to her eighth child and all were girls! Her doctor had assured her that the chance of the eighth child being a girl was less than 1 in 100 . What was the real probability that the eighth child would be a girl? (A) 0.0039 (B) 0.5 (C) \((0.5)^{7}\) (D) \((0.5)^{8}\) (E) \(\frac{(0.5)^{7}+(0.5)^{8}}{2}\)
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