1) Entry to a certain University is determined by a national test. The scores on this
test are normally distributed with a mean of 500 and a standard deviation of 100.
Tom wants to be admitted to this university and he knows that he must score
better than at least 70% of the students who took the test. Tom takes the test and
scores 585. Will he be admitted to this university?
^ I already did this question and I got 80.23%
but I need help with the other questions that go with this :
2) For the same test, consider Sarah, whose score is 683. What is her
a) Z score
b) T score
c) Percentile rank
d) What percentage of people scored between Sarah and the mean? How many
people were ahead of her? (Clue: use the Z table).
e) If the mean on the test were 550 and the Sd = 50, what would be Tom and
Sarah’s scores? Would either of them qualify for the University now?
I did not check #1.
2. Z = (score-mean)/SD
Z = (683-500)/100 (calculate)
T = 50 + 10Z
Percentile rank = proportion ≤ score
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions related to the Z scores for c, d and e.
2) For Sarah's score of 683, we can calculate the following:
a) Z score:
The Z score represents the number of standard deviations that a data point is away from the mean. It is calculated using the formula:
Z = (X - μ) / σ
Here, X is the score, μ is the mean, and σ is the standard deviation.
For Sarah:
Z = (683 - 500) / 100 = 1.83
b) T score:
To calculate the T score, we need to know the degrees of freedom (df), which is usually the sample size minus 1. Since the degrees of freedom are not provided in the question, it's hard to calculate the T score accurately. However, in practice, the T score is often not used for normal distributions, so you can skip this part.
c) Percentile rank:
The percentile rank represents the percentage of scores that fall below a given score. It is calculated using the formula:
Percentile rank = (number of scores below the given score / total number of scores) * 100
To find the percentile rank, we need to find the area to the left of the score on the standard normal distribution. Using the Z score table or a statistical calculator, we can determine that the area to the left of Z = 1.83 is approximately 0.9664.
Therefore, the percentile rank for Sarah's score of 683 is:
Percentile rank = 0.9664 * 100 = 96.64%
d) Percentage of people between Sarah and the mean:
To find the percentage of people who scored between Sarah and the mean, we need to find the area between the Z score of Sarah's score and the Z score of the mean. In this case:
Z_mean = (500 - 500) / 100 = 0
Using the Z table, we can find the area to the left of Z = 0 is 0.5. Therefore, the percentage of people who scored between Sarah and the mean is:
Area_between = 0.9664 - 0.5 = 0.4664 or 46.64%
To find the number of people who scored ahead of Sarah, we can subtract the percentage between Sarah and the mean from 100%:
People_ahead = 100% - 46.64% = 53.36%
e) If the mean on the test were 550 and the standard deviation (SD) was 50, we can calculate the scores for Tom and Sarah using the Z score formula:
For Tom:
Z = (585 - 550) / 50 = 0.7
Tom's score = Z * SD + mean = 0.7 * 50 + 550 = 585
For Sarah:
Z = (683 - 550) / 50 = 2.66
Sarah's score = Z * SD + mean = 2.66 * 50 + 550 = 683
Comparing the new scores with the previous requirement, Tom's score remains unchanged, so he would still qualify for the university. Sarah's score also remains the same, so she would still qualify as well.
To answer these questions, we need to understand some basic concepts related to normal distribution, z-scores, t-scores, and percentiles.
1) To find out if Tom will be admitted to the university, we need to determine his percentile rank. The percentile rank represents the percentage of scores that are equal to or below a particular score.
To find Tom's percentile rank:
Step 1: Calculate the z-score for Tom's score, using the formula: Z = (X - μ) / σ, where X is Tom's score, μ is the mean, and σ is the standard deviation.
Z = (585 - 500) / 100 = 0.85
Step 2: Find the area to the left of Tom's z-score from the standard normal distribution table. This area represents the percentage of scores that are equal to or below Tom's score.
From the z-table, the area to the left of Z = 0.85 is approximately 0.8023, which is 80.23%.
Since Tom's score is better than 80.23% of the students who took the test, he will be admitted to the university.
Now, let's move on to the questions related to Sarah.
2a) To find Sarah's z-score:
Z = (X - μ) / σ
Z = (683 - 500) / 100 = 1.83
2b) T-scores are used when the sample size is small. We don't have the information about the sample size, so we cannot determine the t-score for Sarah.
2c) Percentile rank represents the percentage of scores that are equal to or below a particular score. To find Sarah's percentile rank, we need to find the area to the left of her z-score using the z-table.
From the z-table, the area to the left of Z = 1.83 is approximately 0.9664, which is 96.64%.
So, Sarah's percentile rank is 96.64%, indicating that her score is better than 96.64% of the students who took the test.
2d) To find the percentage of people who scored between Sarah and the mean, we need to calculate the area between the z-scores of Sarah and the mean using the z-table.
From the z-table, the area to the left of Z = 0 (mean) is 0.5, and the area to the left of Z = 1.83 is approximately 0.9664. Therefore, the area between the mean and Sarah's z-score is 0.9664 - 0.5 = 0.4664, which is 46.64%.
To find the number of people ahead of Sarah, we subtract the percentage of people in between Sarah and the mean from 100%. So, the percentage of people ahead of Sarah is 100% - 46.64% = 53.36%.
2e) If the mean on the test were 550 and the standard deviation were 50, we can recalculate the scores for Tom and Sarah using the same formulas as before.
For Tom:
Z = (X - μ) / σ = (585 - 550) / 50 = 0.70
Using the z-table, the area to the left of Z = 0.70 is approximately 0.7580, which is 75.80%.
Since Tom's score is better than 75.80% of the students, he would still qualify for the university with these new parameters.
For Sarah:
Z = (X - μ) / σ = (683 - 550) / 50 = 2.66
Using the z-table, the area to the left of Z = 2.66 is approximately 0.9962, which is 99.62%.
Sarah's score would also make her qualify for the university with these new parameters.
Please note that these calculations are based on the assumption of a normal distribution and may vary slightly depending on the accuracy of the z-table used.