Humber's Financial Services Program is offering a post-diploma program in Securities and Investments. Students must write an admission test before acceptance into the program. The scores on the admission test are normally distributed with a mean score of 150 and a standard deviation of 15.

If admission to the program is based entirely on the score on the admission test and only the top 15% will be accepted, find the cutoff score for gaining admission to the Securities and Investments Program. Round your answer to 2 decimal places.

__________

Given a mean score of 150 and standard deviation of 15, the cut-off score for admission into the Securities and Investment Program, representing the top 15% of scores, is approximately 165.

Explanation:

To answer this question, we need to understand how normal distributions work. In a normal distribution, the mean score is considered the middle of the distribution, with approximately 50% of scores being above this mean, and 50% below it. Standard deviation measures how spread out the distribution is.

In this case, we are looking for a score that is higher than 85% of all scores (since we want the top 15% that will be accepted). For normally distributed data, the score that is 1 standard deviation above the mean encompasses roughly the top 15% of scores.

Given the information in the question, the mean is 150 and the standard deviation is 15, so we calculate: Mean plus one standard deviation = 150 + 15 = 165

Therefore, to be in the running for acceptance into the Post-Diploma program, a student must score at least 165 on the admissions test, rounded off to 2 decimal places.

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MadsMikkelsen MadsMikkelsen · Answer 2 · 2023-11-19T18:10:08-05:00

Final answer:

The cutoff score for gaining admission to the Securities and Investment Program at Humber's Financial Services is approximately 165.54, which represents the 85th percentile in the normal distribution of scores.

Explanation:

Understanding percentiles, standard deviation, and normal distribution is crucial to answer this question.

Firstly, since only the top 15% of scores qualify for admission, this means we are looking for the 85th percentile (100% - 15% = 85%) in the distribution of scores. In a normal distribution, the 85th percentile corresponds approximately to a z-score of 1.036

A z-score represents how many standard deviations a given score is above or below the mean. We can use the formula for converting a z-score to an original score (X):

X = μ + Zσ

where:

μ is the mean
Z is the z-score
σ is the standard deviation.

In this case, μ is 150, Z is 1.036, and σ is 15.

So, to get the cut-off score, plug the values into the formula: X = 150 + 1.036*15 ≈ 165.54.

Therefore, the cutoff score for gaining admission to the Securities and Investment Program at Humber's Financial Services is approximately 165.54, when rounded to two decimal places.

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