If the student ratings of the overall effectiveness of professors is normally distributed with a mean of 5 and a standard deviation of 1.75, what proportion of the professors will be expected to be rated below 2.55?
This is a z-score
z=(x-mu)/(standard deviation)
z=(2.55-5)/(1.75)
find z then look it up on your "areas under the curve normal distribution table" in the back of your textbook : )
To find the proportion of professors expected to be rated below a certain rating, we need to calculate the cumulative probability of that rating based on the given mean and standard deviation.
In this case, we want to find the cumulative probability of rating below 2.55.
To do this, we can use the standard normal distribution table or a calculator that provides the cumulative distribution function (CDF) of the standard normal distribution.
The formula for standardizing a value from a normal distribution is:
Z = (X - μ) / σ
Where:
Z is the standardized value (also known as the z-score)
X is the rating we want to find the proportion for (2.55 in this case)
μ is the mean (5 in this case)
σ is the standard deviation (1.75 in this case)
Plugging in the values, we get:
Z = (2.55 - 5) / 1.75
= -2.45 / 1.75
≈ -1.4
Now, we need to find the cumulative probability associated with this z-score of -1.4.
Using a standard normal distribution table, we can find that the cumulative probability corresponding to -1.4 is approximately 0.0808.
Therefore, approximately 0.0808 or 8.08% of the professors can be expected to be rated below 2.55.