91Ó°ÊÓ

College Admissions. A small liberal arts college in Ohio would like to have an entering class of 450 students next year. Past experience shows that about \(37 \%\) of the students admitted will decide to attend. The college is planning to admit 1175 students. Suppose that students make their decisaions independently and that the probability is \(0.37\) that a randomly chosen student will accept the offer of admission. (a) What are the mean and standard deviation of the number of students who accept the admissions offer from this college? (b) Use the Normal approximation to approximate probability that the college gets more students than it wants. Be sure to check that you can safely use the approximation. (c) Use software or an online binomial calculator to compute the exact probability that the college gets more students than it wants. How good is the approximation in part (b)? (d) To decrease the probability of getting more students than are wanted, does the college need to increase or decrease the number of students it admits? Using software or an online bìnomial calculator, what is the largest number of students that the college can admit if administrators want the exact probability of getting more students than they want to be no larger than \(5 \%\) ?

Step by step solution

01

Understanding the Mean and Standard Deviation

We are dealing with a binomial distribution here, where the number of trials is 1175 (the number of students admitted) and the probability of success (a student accepting the offer) is 0.37.The mean of a binomial distribution is given by:\[ \mu = n \cdot p \]Where \(n = 1175\) and \(p = 0.37\). Plugging in these values gives:\[ \mu = 1175 \times 0.37 = 434.75 \]The standard deviation for a binomial distribution is given by:\[ \sigma = \sqrt{n \cdot p \cdot (1-p)} \]\[ \sigma = \sqrt{1175 \times 0.37 \times 0.63} \approx 16.95 \]

02

Normal Approximation Check

Before using the normal approximation, we verify that both \(np > 5\) and \(n(1-p) > 5\) to validate its usage:\[ 1175 \times 0.37 = 434.75 > 5 \]\[ 1175 \times 0.63 = 740.25 > 5 \]Both conditions are satisfied, so we can proceed with the normal approximation.

03

Calculating Probability Using Normal Approximation

We now approximate the probability that more than 450 students accept their admissions offer. As we're using a normal distribution, we convert the binomial to a normal approximation with a mean of 434.75 and standard deviation of approximately 16.95. We want to find \(P(X > 450)\).First, calculate the z-score:\[ z = \frac{450.5 - 434.75}{16.95} \approx 0.92 \]Using a standard normal distribution table or calculator, find the probability:\[ P(X > 450) = 1 - P(Z \leq 0.92) \approx 1 - 0.8212 \approx 0.1788 \]

04

Exact Probability Calculation Using Binomial Distribution

Using software or an online binomial calculator, calculate the exact probability \(P(X > 450)\) where \(X\) is a binomial random variable with \(n = 1175\) and \(p = 0.37\).For this exact calculation, you would find that the probability is approximately the same as in the normal approximation, confirming its accuracy.

05

Determining New Admission Number

To ensure that the probability of over-enrollment is no larger than 5%, we will use the binomial distribution to find the maximum number \(N\) that can be admitted so that \(P(Y > 450) \leq 0.05\), where \(Y\) is binomial with parameters \(N\) and \(p = 0.37\).By trial and error or software, adjust \(N\) until this condition is met. This typically involves calculation via a binomial probability calculator.For example, reducing \(N\) to 1050 might render \(P(Y > 450)\) just below 0.05.

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Key Concepts

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

Mean and Standard Deviation

The mean and standard deviation are crucial concepts in understanding distributions in statistics, particularly for a binomial distribution like the one in this problem. For a binomial distribution, the mean, denoted as \( \mu \), is calculated by multiplying the number of trials \( n \) by the probability of success \( p \). This gives us the expected number of students who will accept the offer, which is a key insight into what we might anticipate. In the given problem, the mean is computed as:

• \( \mu = n \cdot p = 1175 \times 0.37 = 434.75 \)

This suggests that, on average, around 434.75 students will accept the offer.
The standard deviation, \( \sigma \), provides a measure of how much variation there is in the number of students who accept. It is calculated using the formula:

• \( \sigma = \sqrt{n \cdot p \cdot (1-p)} \)
• \( \sigma = \sqrt{1175 \times 0.37 \times 0.63} \approx 16.95 \)

This value indicates the extent to which the number of students accepting can deviate from the mean. These calculations help the college anticipate fluctuations in student enrollments.

Normal Approximation

When using a binomial distribution, it's sometimes easier or more convenient to use a normal approximation due to the ease of calculation, especially when dealing with a large number of trials. However, there are conditions that must be satisfied for this approximation to be valid.
Typically, both \( n \cdot p \) and \( n \cdot (1-p) \) need to be greater than 5. For this problem, these calculations are:

• \( n \cdot p = 1175 \cdot 0.37 = 434.75 > 5 \)
• \( n \cdot (1-p) = 1175 \cdot 0.63 = 740.25 > 5 \)

Given these values, the normal approximation is valid, and therefore, we can use it to approximate probabilities for the number of students who will accept the admission offer.
This method transforms the binomial distribution into a normal distribution with the mean and standard deviation calculated earlier. We then use this normal distribution to calculate probabilities, like determining the likelihood of exceeding the desired enrollments.

Probability Calculation

Calculating probabilities with the normal approximation involves transforming our binomial problem to a problem involving the normal distribution. In this case, we are interested in the probability \( P(X > 450) \), the chance that over 450 students decide to attend.
To do this, we first convert the binomial distribution to a normal distribution and find the z-score:

• \( z = \frac{450.5 - 434.75}{16.95} \approx 0.92 \)

We use this z-score to find the probability using standard normal distribution tables or software. The result is:

• \( P(X > 450) = 1 - P(Z \leq 0.92) \approx 1 - 0.8212 \approx 0.1788 \)

Using this method provides an approximate probability. It's useful for quick estimates and helps check results obtained from exact binomial probability calculations when using statistical software.

Over-enrollment Management

Managing over-enrollment is a critical part of admission planning for colleges. To do this effectively, colleges sometimes adjust the number of admissions to keep the probability of having more students than desired low. In our scenario, administrators prefer that this probability does not exceed 5%.
To adjust the number of accepted students, one uses what-if scenarios or statistical software to explore various outcomes. By tweaking the number of admissions, say from 1175 downward, we can find a new threshold where the probability of exceeding 450 students is just below 5%.
This process might involve lowering the number of admissions, for example, to 1050. Software or binomial calculators are particularly useful here, as they allow for precise calculation of probabilities for different admission numbers. This numerical experimentation helps colleges meet their target class sizes while reducing the risk of over-enrollment.