For a class of 43 students fora certain exam, the mean grade is 67 and the standard deviation is 7.
a) How many students should receive a C? (a c is between 74-79)
b) How many students should receive a B? (a B is between a 80-89)
c) what score is necessary to get an A?
use this applet
http://davidmlane.com/hyperstat/z_table.html
enter your mean and SD
then click on the "between" and enter 74 and 79
multiply the probability factor by 43
To answer these questions, we need to use the concept of the normal distribution and the z-score.
a) To find out how many students should receive a C (between 74-79), we need to find the percentage of students within this range.
First, we calculate the z-scores for both ends of the range:
Z1 = (74 - 67) / 7 = 1
Z2 = (79 - 67) / 7 = 1.71
Next, we find the area under the normal curve between these two z-scores. We can use a standard normal distribution table or a calculator, or use a z-table tool online.
From the standard normal distribution table or calculator, we find that the area between z = 1 and z = 1.71 is approximately 0.219.
Finally, we multiply the percentage by the total number of students in the class:
0.219 * 43 = 9.417
Therefore, approximately 9 students should receive a C.
b) To find out how many students should receive a B (between 80-89), we follow the same steps:
Z1 = (80 - 67) / 7 = 1.857
Z2 = (89 - 67) / 7 = 3.143
From the standard normal distribution table or calculator, we find that the area between z = 1.857 and z = 3.143 is approximately 0.239.
0.239 * 43 = 10.277
Therefore, approximately 10 students should receive a B.
c) To determine the score necessary to get an A, we need to find the z-score that corresponds to the desired percentage.
Since an A is typically the top percentage, we want to find the z-score that corresponds to, for example, the top 10% of students. This leaves 90% of the distribution below it.
Using a z-table, we find the z-score that corresponds to a cumulative area of 0.90, which is approximately 1.28.
Next, we use the formula:
Z = (X - mean) / standard deviation
Rearranging the formula to solve for X, where X is the score needed for an A:
X = (Z * standard deviation) + mean
X = (1.28 * 7) + 67
X ≈ 75.96
Therefore, a score of approximately 75.96 or higher is necessary to get an A.
To answer these questions, we need to make use of the concept of z-scores. A z-score tells us how many standard deviations a particular value is away from the mean.
a) To calculate the number of students who should receive a C (between 74-79), we need to find the z-scores corresponding to these values and then determine the proportion of students within this range.
First, let's calculate the z-scores:
Lower bound: (74 - 67) / 7 = 1
Upper bound: (79 - 67) / 7 = 12/7 ≈ 1.71
Next, we need to find the proportion of students between these z-scores. We can use a standard normal distribution table (also known as the z-table) to determine this.
Using the z-table, the proportion of students between the lower and upper bound z-scores approximately corresponds to the area under the standard normal curve between those z-scores.
By consulting the z-table, we find that the area to the left of a z-score of 1 is approximately 0.8413, and the area to the left of a z-score of 1.71 is approximately 0.9554. So, the proportion of students between these z-scores is approximately 0.9554 - 0.8413 = 0.1141.
Now, multiply this proportion by the total number of students:
0.1141 * 43 ≈ 4.9
Therefore, approximately 4 or 5 students should receive a C.
b) Follow the same process as above to find the number of students who should receive a B (between 80-89).
Lower bound z-score: (80 - 67) / 7 ≈ 1.8571
Upper bound z-score: (89 - 67) / 7 ≈ 3.1429
Again, consulting the z-table, we find the area to the left of a z-score of 1.8571 is approximately 0.9643, and the area to the left of a z-score of 3.1429 is approximately 0.9994. So, the proportion of students between these z-scores is approximately 0.9994 - 0.9643 = 0.0351.
Multiply this proportion by the total number of students:
0.0351 * 43 ≈ 1.51
Therefore, approximately 1 or 2 students should receive a B.
c) To determine the score necessary to get an A, we need to find the corresponding z-score for the desired area under the standard normal curve.
An A corresponds to a z-score that is greater than a certain value, which we will denote as z_A. The area under the standard normal curve to the left of z_A is equal to 1 - (proportion of students receiving a B + proportion of students receiving a C).
Thus, the desired area for an A is approximately 1 - (0.1141 + 0.0351) = 0.8508.
To find the corresponding z-score for this area, we can use the inverse of the cumulative distribution function (CDF) of the standard normal distribution. Alternatively, we can use a z-table to find the z-score that corresponds to an area of 0.8508.
By consulting the z-table, we find that the z-score corresponding to an area of 0.8508 is approximately 1.0364.
Now, we can use the formula:
z_A = (score - mean) / standard deviation
Rearranging the formula to solve for the score:
score = (z_A * standard deviation) + mean
Plugging in the values:
score = (1.0364 * 7) + 67
Therefore, the score necessary to get an A is approximately 74.25 (rounded to the nearest hundredth).