1) A scale measuring prejudice has been administered to a large sample of respondents. The distribution of scores is approximately normal with a mean of 31 and a standard deviation of 5. What percentage of sample had scores below 20?
normalcdf(-100,20,30,5)
0.0228
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.
To find the percentage of the sample with scores below 20, we need to calculate the z-score for a score of 20, and then use the standardized normal distribution table (also known as the Z-table) to find the corresponding percentage.
First, let's calculate the z-score:
z = (x - mean) / standard deviation
where x is the score we want to find the percentage for, mean is the mean of the distribution, and standard deviation is the standard deviation of the distribution.
In this case:
x = 20
mean = 31
standard deviation = 5
z = (20 - 31) / 5
z = -2.2
Now, we can look up the percentage associated with a z-score of -2.2 in the standardized normal distribution table. The Z-table provides the percentage of values that fall below a given z-score.
The Z-table typically shows decimal values between 0 and 1, so we need to convert our z-score to a positive value to find the corresponding percentage.
Looking up a z-score of 2.2 gives a percentage of 0.0139.
However, since we are interested in the percentage of scores below 20, which is to the left of the mean, we need to subtract this percentage from 1 to get the percentage below 20.
Percentage below 20 = 1 - 0.0139 = 0.9861
So, approximately 98.61% of the sample had scores below 20.