4. A large company insists that all job applicants who are invited for an interview take a psychometric test. The results of these tests follow a normal distribution with a mean of 61% and a standard deviation of 7.2%.
a) What proportion of applicants would be expected to score over 70%?
b) What proportion of applicants would be expected to score under 40%?
c) What proportion of applicants would be expected to score between 50% and 65%?
Use same process as indicated in your previous post.
To answer these questions, we need to use the concept of standard deviation and the normal distribution.
The formula to calculate the proportion of values within a specific range in a normal distribution is:
P(a < X < b) = P(X < b) - P(X < a)
where P represents the probability, X represents the variable in question, and a and b represent the lower and upper bounds of the range.
Now let's calculate the answers to each question:
a) To find the proportion of applicants who would be expected to score over 70%, we need to calculate P(X > 70).
First, we need to convert the raw score of 70 to a standardized score (z-score) by subtracting the mean (61%) and then dividing the result by the standard deviation (7.2%):
z = (70 - 61) / 7.2 = 1.25
Using a standard normal distribution table or a statistical calculator, we can find the proportion of values to the right of the z-score of 1.25.
P(X > 70) = P(Z > 1.25) ≈ 1 - 0.8944 ≈ 0.1056
Therefore, approximately 10.56% of applicants would be expected to score over 70%.
b) To find the proportion of applicants who would be expected to score under 40%, we need to calculate P(X < 40).
Following the same steps as in question a), we find the z-score for 40%:
z = (40 - 61) / 7.2 = -2.92
Using a standard normal distribution table or a statistical calculator, we can find the proportion of values to the left of the z-score of -2.92.
P(X < 40) = P(Z < -2.92) ≈ 0.0019
Therefore, approximately 0.19% of applicants would be expected to score under 40%.
c) To find the proportion of applicants who would be expected to score between 50% and 65%, we need to calculate P(50 < X < 65).
Again, following the same steps as before, we find the z-scores for both 50% and 65%:
z1 = (50 - 61) / 7.2 = -1.53
z2 = (65 - 61) / 7.2 = 0.56
Using a standard normal distribution table or a statistical calculator, we can find the proportions of values to the left of these z-scores.
P(50 < X < 65) = P(-1.53 < Z < 0.56) ≈ 0.6255 - 0.0630 ≈ 0.5625
Therefore, approximately 56.25% of applicants would be expected to score between 50% and 65%.