After years of data collection by college students, it has been determined that the distribution of course grades in Ms. Green’s Statistic II class follows a normal distribution with a mean of 62 and a standard deviation of 14.
(a) A student will pass Ms. Green’s course if he or she receives a course grade of 70 or higher. A current student in Ms Green’s class is chosen at random. What is the probability that the student will NOT receive a passing grade in Ms. Green’s class?
(b) What is the probability of a student receiving a “B” (i.e. a course grade greater than or equal to 80, but less than 90) in Ms. Green’s class?
(c) What range of course grade values would be considered outliers for Ms. Green’s class?
a. Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to that Z score.
b. Use same equation to find proportions between the two Z scores.
c. I would say any score > ± 3 SD.
I'll let you do the calculations.
To answer these questions, we need to use the properties of the normal distribution. We'll use the mean and standard deviation provided to calculate probabilities and ranges.
(a) To find the probability that a student will NOT receive a passing grade, we need to calculate the area under the normal distribution curve to the left of 70.
To do this, we can use a standard normal distribution table or calculator and convert the value of 70 into a z-score. The z-score is the number of standard deviations away from the mean.
Using the formula z = (x - μ) / σ, where x is the value we want to convert, μ is the mean, and σ is the standard deviation, we can calculate the z-score for 70:
z = (70 - 62) / 14 = 0.5714
Finding the area corresponding to this z-score, we can say that the probability of getting a grade less than 70 is equal to the area to the left of the z-score 0.5714.
(b) To find the probability of a student receiving a "B", we need to calculate the area under the normal distribution curve between 80 and 90.
First, we calculate the z-scores for both 80 and 90.
For 80:
z1 = (80 - 62) / 14 = 1.2857
For 90:
z2 = (90 - 62) / 14 = 2.0
Next, we find the area between these two z-scores. This can be done by subtracting the area to the left of z2 from the area to the left of z1.
P(80 ≤ X < 90) = P(z1 ≤ Z < z2)
Now, using a standard normal distribution table or calculator, we can find the probabilities corresponding to these z-scores and subtract them.
(c) To determine the range of course grade values considered outliers, we need to define what we mean by outliers. A common way to identify outliers in a normal distribution is by using the concept of standard deviation.
Using the empirical rule for normally distributed data, we can say that approximately 68% of the data falls within 1 standard deviation of the mean, about 95% falls within 2 standard deviations, and approximately 99.7% falls within 3 standard deviations.
In Ms. Green's class, we can consider course grade values outside of 3 standard deviations from the mean to be outliers. So we need to determine the range of values that are more than 3 standard deviations below the mean and more than 3 standard deviations above the mean.
To find these values, we can calculate:
Lower Outlier Value = Mean - (3 * Standard Deviation)
Upper Outlier Value = Mean + (3 * Standard Deviation)
This range would be considered as outliers in Ms. Green's class.