Question 55% Of Students Applying To A University
Question 1fifty Five Percent Of The Students Applying To A University
Question 1 Fifty-five percent of the students applying to a university are accepted. Using the binomial probability tables or Excel, what is the probability that among the next 18 applicants: Exactly 6 will be accepted? At least 10 will be accepted? Exactly 5 will be rejected? Fifteen or more will be accepted? Determine the expected number of acceptances. Compute the standard deviation.
Scores on a recent national statistics exam were normally distributed with a mean of 82 and a standard deviation of 4. What is the probability that a randomly selected exam will have a score of at least 87? What percentage of exams will have scores between 80 and 85? If the top 3% of test scores receive merit awards, what is the lowest score eligible for an award?
Paper For Above instruction
The problems presented involve applications of binomial probability theory and normal distribution concepts, both fundamental in statistical analysis. The first part deals with binomial probabilities related to university admissions, while the second focuses on properties of the normal distribution concerning exam scores.
Part 1: Binomial Probability Calculations for University Acceptance
Given that 55% of applicants are accepted, we define the probability of acceptance (success) as p = 0.55. The random variable X, representing the number of accepted students among the next 18 applicants, follows a binomial distribution: X ~ Binomial(n=18, p=0.55).
- Probability that exactly 6 will be accepted:
Using the binomial probability formula:
\[ P(X=6) = \binom{18}{6} (0.55)^6 (0.45)^{12} \]
Alternatively, Excel’s BINOM.DIST function or binomial tables can be used:
```excel
=BINOM.DIST(6,18,0.55,FALSE)
```
- Probability that at least 10 will be accepted:
Calculates as the complement of having fewer than 10 accepted:
```excel
=1 - BINOM.DIST(9,18,0.55,TRUE)
```
- Probability that exactly 5 will be rejected:
Rejections are 1 - acceptance, so rejecting exactly 5 students translates to acceptance of 13 students:
```excel
P(X=13) = BINOM.DIST(13,18,0.55,FALSE)
```
- Probability that 15 or more will be accepted:
Sum probabilities for 15, 16, 17, and 18 accepted students:
```excel
=1 - BINOM.DIST(14,18,0.55,TRUE)
```
- Expected number of acceptances and standard deviation:
The expected value (mean) μ = n p = 18 0.55 = 9.9.
The standard deviation σ = \(\sqrt{n p (1 - p)}\) = \(\sqrt{18 0.55 0.45}\) ≈ 2.52.
Part 2: Normal Distribution of Exam Scores
Given the mean (\(\mu\)) = 82 and standard deviation (\(\sigma\)) = 4 for exam scores, we analyze probabilities for specific score thresholds using standard normal distribution (Z-scores).
- Probability that a score is at least 87:
Calculate Z for score 87:
\[ Z = \frac{87 - 82}{4} = 1.25 \]
Using standard normal tables or Excel's NORM.DIST:
```excel
=1 - NORM.DIST(87,82,4,TRUE)
```
This yields approximately 0.1056, or 10.56%.
- Percentage of scores between 80 and 85:
Calculate Z-scores:
\[ Z_{80} = \frac{80 - 82}{4} = -0.5 \]
\[ Z_{85} = \frac{85 - 82}{4} = 0.75 \]
Using cumulative distribution:
```excel
=NORM.DIST(85,82,4,TRUE) - NORM.DIST(80,82,4,TRUE)
```
This result approximates 0.5328, meaning about 53.28% of scores fall between 80 and 85.
- Lowest score for top 3% of scores:
Find Z-value corresponding to the 97th percentile:
```excel
=NORM.INV(0.97,82,4)
```
The Z-score for the 97th percentile is approximately 1.88, so the lowest score:
\[ 82 + (1.88 * 4) \approx 89.52 \]
Therefore, scores above approximately 89.52 qualify for the merit award.
These calculations demonstrate the application of binomial and normal distribution formulas to real-world data, highlighting the importance of these tools in decision-making and statistical inference.
References:
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- Moore, D. S., McCabe, G. P., & Craig, B. A. (2012). Introduction to the Practice of Statistics. W. H. Freeman.
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- Wilcox, R. R. (2012). Introduction to Robust Estimation and Hypothesis Testing. Academic Press.
- Johnson, N. L., Kotz, S., & Balakrishnan, N. (1994). Continuous Univariate Distributions. Wiley.
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