Mr. Smith has 100 students in his Algebra classes. The grades of a quiz are normally distributed with a mean of 70 and a standard deviation of 4. Sally claims that 75 students scored between 66 and 74 on this quiz. Given that this is normally distributed, is Sally correct?
I thought I already answered this.
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability between the two Z scores. Is it .75?
To determine if Sally's claim is correct, we need to calculate the probability of having 75 students score between 66 and 74 on the quiz, assuming a normal distribution with a mean of 70 and a standard deviation of 4.
To do this, we'll convert the scores into z-scores by using the z-score formula:
z = (x - μ) / σ
Where:
- x is the observed value (score)
- μ is the mean
- σ is the standard deviation
For 66:
z = (66 - 70) / 4
z = -4 / 4
z = -1
For 74:
z = (74 - 70) / 4
z = 4 / 4
z = 1
We can use a standard normal distribution table or a calculator to find the probability associated with a z-score of -1 and a z-score of 1. We want to find the probability of students scoring between these z-scores.
P(-1 < z < 1) = P(z < 1) - P(z < -1)
Using a standard normal distribution table, we find that P(z < 1) is approximately 0.8413 and P(z < -1) is approximately 0.1587.
P(-1 < z < 1) = 0.8413 - 0.1587
P(-1 < z < 1) = 0.6826
Therefore, the probability of having a student score between 66 and 74 on the quiz is approximately 0.6826. Now, we need to check if this probability supports Sally's claim.
Sally claims that 75 students scored between 66 and 74. We can calculate the expected number of students scoring between 66 and 74 by multiplying the probability by the total number of students.
Expected number = Probability * Total number of students
Expected number = 0.6826 * 100
Expected number = 68.26
Since the expected number of students scoring between 66 and 74 is 68.26 and Sally claims 75 students, her claim seems reasonable. However, we should note that the calculation assumes a perfect normal distribution and may not be exactly accurate in practice.