14.
The marks on a statistics test are normally distributed with a mean of 62 and a variance of 225. If the instructor wishes to assign B’s or higher to the top 30% of the students in the class, what mark is required to get a B or higher?
a. 68.7 b. 71.5 c. 73.2 d. 74.6 e.69.9
The answer is E, but don't know why.
2. Suppose the test scores of 600 students are normally distributed with a mean of 76 and variance
of 64. The number of students scoring between 70 and 82 is:
a. 272 b. 164 c. 260 d. 136 e. 328
The answer is E but don't know why.
14. Variance = SD^2
Z = (score-mean)/SD = (score-62)/15
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion (.30) to give you the Z score. Put that value in the above equation to find the score.
Z = ([70 or 82]-76)/8
Use the Z scores in the same table to find the proportions between them and multiply by 600.
Well, let me try to answer these questions using my unique brand of humor!
For the first question, we know that the marks are normally distributed, so we can use the z-score formula to find the mark required for a B or higher. The z-score formula is z = (x - mean) / standard deviation. In this case, the mean is 62 and the variance is 225, so the standard deviation is the square root of the variance, which is 15.
Since the instructor wants to assign B's or higher to the top 30% of the students, we need to find the z-score that corresponds to the top 30%. To find this, we look up the z-score in the z-table, and we find that the z-score is approximately 0.524 (rounded to three decimal places).
Now we can use the z-score formula to solve for x: z = (x - mean) / standard deviation. Plugging in the values we know, we get 0.524 = (x - 62) / 15. Solving this equation gives us x = 69.86, which is closest to option E, 69.9.
For the second question, we want to find the number of students scoring between 70 and 82. Since the scores are normally distributed with a mean of 76 and a standard deviation of 8 (sqrt(64)), we can calculate the z-scores for 70 and 82 using the z-score formula.
For 70, the z-score is (70 - 76) / 8 = -0.75, and for 82, the z-score is (82 - 76) / 8 = 0.75. Now we look up these z-scores in the z-table and subtract the cumulative probability to find the percentage of students between these scores.
The cumulative probability for -0.75 is approximately 0.2261, and the cumulative probability for 0.75 is approximately 0.7739. To find the percentage in between, we subtract 0.2261 from 0.7739, which gives us 0.5478 or 54.78%.
Finally, we multiply this percentage by the total number of students, which is 600, to find the number of students scoring between 70 and 82. 54.78% of 600 is approximately 328, which is closest to option E.
So, there you have it! I hope my humor didn't take away from the explanation too much. If you have any more questions, feel free to ask!
To calculate the mark required to get a B or higher, we need to find the z-score corresponding to the top 30% of the distribution.
Step 1: Convert the top 30% to a z-score
To find the z-score, we look up the corresponding percentile in the standard normal distribution table.
The percentile corresponding to the top 30% is 1 - 0.30 = 0.70.
Using the standard normal distribution table or a calculator, we find that the z-score corresponding to a percentile of 0.70 is approximately 0.52.
Step 2: Use the z-score to find the mark
The z-score formula is:
z = (x - mean) / sqrt(variance)
Rearranging the formula to solve for x (mark):
x = z * sqrt(variance) + mean
Substituting the given values into the formula:
x = 0.52 * sqrt(225) + 62
x = 0.52 * 15 + 62
x = 7.8 + 62
x ≈ 69.9
Therefore, a mark of approximately 69.9 or higher is required to get a B or higher.
For the second question:
Step 1: Convert the given scores to z-scores
To find the z-scores for scores of 70 and 82, we use the z-score formula:
z = (x - mean) / sqrt(variance)
For 70:
z_70 = (70 - 76) / sqrt(64)
z_70 = -6 / 8
z_70 = -0.75
For 82:
z_82 = (82 - 76) / sqrt(64)
z_82 = 6 / 8
z_82 = 0.75
Step 2: Calculate the area under the curve between the z-scores
To find the number of students scoring between 70 and 82, we need to calculate the area under the normal distribution curve between the z-scores of -0.75 and 0.75.
Using the standard normal distribution table or a calculator, we can find that the area under the curve between -0.75 and 0.75 is approximately 0.548.
Step 3: Calculate the number of students
Since the distribution is approximately normal, we can estimate the number of students within this range by multiplying the total number of students (600) by the proportion of students within this range (0.548):
Number of students = 600 * 0.548 ≈ 328
Therefore, the number of students scoring between 70 and 82 is approximately 328.
To find the mark required to get a B or higher, we need to determine the corresponding z-score for the top 30% of the students in the class.
1. For the first question:
a) We know that the mean (μ) is 62 and the variance (σ^2) is 225. To find the standard deviation (σ), we take the square root of the variance: σ = √225 = 15.
b) The z-score formula is z = (x - μ) / σ, where x is the mark we want to find.
c) We need to find the z-score for the top 30% of the distribution. Since the distribution is normal, we can use the z-table or a statistical calculator to find the z-score associated with a cumulative area of 0.70 (which represents the top 30% of scores). The z-score associated with a cumulative area of 0.70 is approximately 0.524.
d) Now, we can plug in the information into the z-score formula and solve for x:
0.524 = (x - 62) / 15
e) Simplifying the equation, we get:
0.524 * 15 = x - 62
7.86 + 62 = x
f) Calculating the result, we find that x ≈ 69.86.
Therefore, the mark required to get a B or higher is approximately 69.9. Therefore, the correct option for the first question is e. 69.9.
2. For the second question:
a) We know that the mean (μ) is 76 and the variance (σ^2) is 64. To find the standard deviation (σ), we take the square root of the variance: σ = √64 = 8.
b) We need to find the number of students scoring between 70 and 82, which corresponds to finding the percentage of students within this range.
c) We can convert the given scores to z-scores and use the z-table or a statistical calculator to find the cumulative area associated with each z-score.
d) The z-score for 70 is calculated as:
z = (70 - 76) / 8 = -6 / 8 = -0.75
e) The z-score for 82 is calculated as:
z = (82 - 76) / 8 = 6 / 8 = 0.75
f) Using the z-table or a statistical calculator, we find that the cumulative area associated with a z-score of -0.75 is approximately 0.2266, and the cumulative area associated with a z-score of 0.75 is approximately 0.7734.
g) The difference between these cumulative areas represents the percentage of students scoring between 70 and 82:
0.7734 - 0.2266 = 0.5468
h) Multiplying this percentage by the total number of students (600) gives us the number of students scoring between 70 and 82:
0.5468 * 600 ≈ 328.08
Since we can't have fractional students, we round the result to the nearest whole number.
Therefore, the number of students scoring between 70 and 82 is approximately 328. Therefore, the correct option for the second question is e. 328.