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Mean and Standard Deviation of Class Year In Exercise P.117, we discuss the random variable counting the number of seniors in a sample of four undergraduate students at a university, given that the proportion of undergraduate students who are seniors is \(0.25 .\) Find the mean and standard deviation of this random variable.

Short Answer

Expert verified
The mean (expected value) of the random variable is 1. The standard deviation of the random variable is \(\sqrt{(4 \times 0.25 \times 0.75)}\).

Step by step solution

01

Identifying parameters of the binomial distribution

First identify the parameters for the binomial distribution problem. In this case, the number of trials, \(n\), is the sample size which is 4 (students) and the probability of success, \(p\), is the proportion of seniors which is 0.25.
02

Calculate the mean

Next, calculate the mean, also known as expected value, using the formula for a binomial distribution, which is \(n \times p\). Substituting the given values, the mean (\(\mu\)) is \(4 \times 0.25 = 1\).
03

Calculate the standard deviation

Now, calculate the standard deviation using its formula in the binomial distribution context, which is \(\sqrt{n \times p \times (1-p)}\). Substituting the given values, the standard deviation (\(\sigma\)) is \(\sqrt{(4 \times 0.25 \times 0.75)}\). Compute this to get the value of the standard deviation.

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Key Concepts

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

Understanding Mean Calculation in a Binomial Distribution
The mean of a binomial distribution, also known as the expected value, provides insight into the average outcome after multiple trials. In any given experiment, this average tells us what we expect to happen on average each time.

For a binomial distribution, the formula to determine the mean is quite straightforward:
  • Mean (\[\mu\]) = number of trials (\[n\]) times the probability of success (\[p\]).
In the context of our problem, we are considering 4 undergraduate students, and 25% of them are seniors. So the probability of selecting a senior is 0.25.

Applying the formula: \[\mu = n \times p = 4 \times 0.25 = 1\]
This result indicates that, on average, 1 out of 4 students is expected to be a senior. Understanding the mean helps us step into the realm of probabilities and predictability in random selections.
Standard Deviation Calculation in a Binomial Distribution
The standard deviation tells us about the variability in the data. In a binomial distribution, it measures how much our data points differ from the mean, providing a sense of how spread out they are.

In binomial distributions, the formula for standard deviation (\[\sigma\]) is:
  • Standard deviation (\[\sigma\]) = \[\sqrt{n \times p \times (1-p)}\].
It factors in:
  • The number of trials (\[n\]).
  • The probability of success (\[p\]).
  • The probability of failure (\[1-p\]).
Using our example, with 4 trials and a 0.25 probability of success, the calculation becomes: \[\sigma = \sqrt{4 \times 0.25 \times 0.75}\]
Perform this calculation to find the standard deviation. This number will show the average distance any one trial is from the mean, offering an understanding of how much diversity exists in the sample outcomes.
Probability of Success in a Binomial Setting
The probability of success in a binomial distribution scenario is a crucial component. It represents the likelihood of each individual trial resulting in what we've defined as "success."

Defined as (\[p\]), the probability of success helps us calculate the mean and standard deviation, anchoring the entire binomial framework.
  • A probability of success (\[p\]) must always fall between 0 and 1, inclusive.
  • In practical terms, it’s the expected ratio or percentage representing success in the data set.
In our exercise, the probability was given as 0.25. This number signifies that in the group of undergraduate students, 25% are seniors. Each time a student is randomly chosen, there is a 25% chance they are in their senior year.

Grasping this concept is key in evaluating scenarios where outcomes can be categorized as success or failure, thanks to its foundation in expectation and variance.

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Most popular questions from this chapter

In Exercises \(\mathrm{P} .74\) to \(\mathrm{P} .77\), fill in the \(?\) to make \(p(x)\) a probability function. If not possible, say so. $$ \begin{array}{lcccc} \hline x & 10 & 20 & 30 & 40 \\ \hline p(x) & 0.2 & 0.2 & ? & 0.2 \\ \hline \end{array} $$

Find endpoint(s) on the given normal density curve with the given property. P.147 (a) The area to the left of the endpoint on a \(N(100,15)\) curve is about 0.75 (b) The area to the right of the endpoint on a \(N(8,1)\) curve is about 0.03 .

Find the specified areas for a normal density. (a) The area below 0.21 on a \(N(0.3,0.04)\) distribution (b) The area above 472 on a \(N(500,25)\) distribution (c) The area between 8 and 10 on a \(N(15,6)\) distribution

In a bag of peanut \(M\) \& M's, there are \(80 \mathrm{M} \& \mathrm{Ms}\), with 11 red ones, 12 orange ones, 20 blue ones, 11 green ones, 18 yellow ones, and 8 brown ones. They are mixed up so that each candy piece is equally likely to be selected if we pick one. (a) If we select one at random, what is the probability that it is red? (b) If we select one at random, what is the probability that it is not blue? (c) If we select one at random, what is the probability that it is red or orange? (d) If we select one at random, then put it back, mix them up well (so the selections are independent) and select another one, what is the probability that both the first and second ones are blue? (e) If we select one, keep it, and then select a second one, what is the probability that the first one is red and the second one is green?

Find the specified areas for a \(N(0,1)\) density. (a) The area above \(z=-2.10\) (b) The area below \(z=-0.5\) (c) The area between \(z=-1.5\) and \(z=0.5\)

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