Answer the following. Show your work to get credit.
Exam scores are normally distributed with a mean of 81 and a standard deviation of 9.
a) What is the minimum score one must have to be in the top 4% of the students taking the exam?
b) What percentage of scores are below 90?
c) What percentage of scores are above 45?
d) If 30 people take this exam, how many would you expect to get above a 93%?
e) If you have a test score in the 45th percentile, what is its z-score?
f) What is the z-score and percentile for an exam score 1.2 standard deviations below the mean?
a) To find the minimum score required to be in the top 4%, we need to find the z-score corresponding to this percentile and then convert it back to the actual score using the formula:
z = (x - mean) / standard deviation
First, we find the z-score using the standard normal distribution table or a statistics calculator. The cumulative probability associated with the top 4% is 1 - 0.04 = 0.96.
Looking up the z-score for a cumulative probability of 0.96, we find it to be approximately 1.75.
Next, we rearrange the formula to solve for x:
x = z * standard deviation + mean
x = 1.75 * 9 + 81
x ≈ 96.75
Therefore, the minimum score one must have to be in the top 4% is approximately 96.75.
b) To find the percentage of scores below 90, we need to find the cumulative probability associated with a z-score of (90 - 81) / 9 = 1.00.
Using the standard normal distribution table or a statistics calculator, we find the cumulative probability for a z-score of 1.00 to be approximately 0.8413.
Converting this probability to a percentage, we multiply by 100:
Percentage = 0.8413 * 100 ≈ 84.13%
Therefore, approximately 84.13% of scores are below 90.
c) Similarly, to find the percentage of scores above 45, we need to find the cumulative probability associated with a z-score of (45 - 81) / 9 = -4.00.
Using the standard normal distribution table or a statistics calculator, we find the cumulative probability for a z-score of -4.00 to be approximately 0.00003.
Converting this probability to a percentage, we multiply by 100:
Percentage = 0.00003 * 100 ≈ 0.003%
Therefore, approximately 0.003% of scores are above 45.
d) To find the number of people expected to get above a 93%, we need to find the cumulative probability associated with a z-score of (93 - 81) / 9 = 1.33.
Using the standard normal distribution table or a statistics calculator, we find the cumulative probability for a z-score of 1.33 to be approximately 0.908.
Therefore, out of 30 people, we expect approximately 0.908 * 30 ≈ 27.24 people to get above a 93%. Since the number of people cannot be fractional, we round to the nearest whole number.
We would expect around 27 people to get above a 93%.
e) To find the z-score for a test score in the 45th percentile, we look up the cumulative probability associated with a z-score.
The percentile can be converted to a cumulative probability by dividing it by 100:
Percentile = 45 / 100 = 0.45
Using the standard normal distribution table or a statistics calculator, we find the z-score for a cumulative probability of 0.45 to be approximately -0.1257.
Therefore, the z-score for a test score in the 45th percentile is approximately -0.1257.
f) To find the z-score and percentile for an exam score 1.2 standard deviations below the mean, we can use the formula:
z = (x - mean) / standard deviation
Substituting the given values, we calculate:
z = (81 - 1.2 * 9 - 81) / 9
z = -1.2
Therefore, the z-score for an exam score 1.2 standard deviations below the mean is -1.2.
To find the corresponding percentile, we look up the cumulative probability associated with a z-score of -1.2 using the standard normal distribution table or a statistics calculator.
The percentile for a z-score of -1.2 is approximately 0.1151, which is equivalent to 11.51%.
Therefore, the exam score 1.2 standard deviations below the mean is at the 11.51th percentile.
To find the answers to these questions, we can use the properties of the normal distribution and z-scores.
a) To find the minimum score required to be in the top 4% of students, we need to find the z-score associated with the 4th percentile. The z-score represents the number of standard deviations a value is from the mean. The z-score formula is given by:
z = (x - μ) / σ
Here, x is the score, μ is the mean, and σ is the standard deviation.
To find the z-score associated with the 4th percentile, we can use a standard normal distribution table or a calculator. The z-score corresponding to the 4th percentile is approximately -1.75.
Using the z-score formula, we can solve for x:
-1.75 = (x - 81) / 9
Solving for x:
-1.75 * 9 = x - 81
-15.75 = x - 81
x = -15.75 + 81
x = 65.25
Therefore, the minimum score one must have to be in the top 4% of students is approximately 65.25.
b) To find the percentage of scores below 90, we need to find the z-score associated with the score 90 and find the corresponding area under the standard normal distribution curve to the left of that z-score.
Using the z-score formula:
z = (x - μ) / σ
z = (90 - 81) / 9
z = 1
The area to the left of a z-score of 1 can be found from a standard normal distribution table, which is approximately 0.8413.
This means that approximately 84.13% of scores are below 90.
c) To find the percentage of scores above 45, we can use the same process as in part b.
Using the z-score formula:
z = (x - μ) / σ
z = (45 - 81) / 9
z = -4
The area to the left of a z-score of -4 is negligible since it is extremely far away from the mean.
However, since the total area under the curve is 1, the percentage above the score of 45 is 1 - the area to the left of z = -4. This is approximately 0%.
Therefore, approximately 0% of scores are above 45.
d) To find the number of people expected to get above a 93%, we can use the z-score formula and the given information.
Using the z-score formula:
z = (x - μ) / σ
z = (93 - 81) / 9
z = 1.3333
The area to the left of a z-score of 1.3333 can be found from a standard normal distribution table or calculator, which is approximately 0.9088.
Therefore, approximately 30 * (1 - 0.9088) = 30 * 0.0912 = 2.736 people are expected to get above a 93%.
Since we can't have a fraction of a person, we would expect around 3 people to get above a 93%.
e) To find the z-score for a test score in the 45th percentile, we can use a standard normal distribution table or a calculator.
The 45th percentile corresponds to a z-score of approximately -0.125.
Therefore, the z-score corresponding to the 45th percentile is approximately -0.125.
f) To find the z-score and percentile for an exam score 1.2 standard deviations below the mean, we can use the z-score formula and the given information.
Using the z-score formula:
z = (x - μ) / σ
z = (81 - 1.2 * 9 - 81) / 9
z = -1.2
The z-score is approximately -1.2.
To find the percentile associated with this z-score, we can use a standard normal distribution table or a calculator. From the table, the area to the left of a z-score of -1.2 is approximately 0.1151.
Therefore, the exam score 1.2 standard deviations below the mean has a percentile of approximately 11.51%.