A standardized test has a mean of 88 and a standard deviation of 12. What is the score at the 90th percentile? Assume a normal distribution.
how would I solve?
a. 90.00
b. 112.00
c. 103.36
d. 91.00
Anyone?!
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 (.90) and its Z score. Insert Z score into equation above.
.3159=(88-?)/12
For .90, Z value should be a number > +1.
Z = (score-88)/12
To find the score at the 90th percentile, you can use the Z-score formula and then convert it back to the original scale using the mean and standard deviation provided.
First, we need to find the Z-score for the 90th percentile. The Z-score corresponds to the number of standard deviations away from the mean a particular value is.
The formula to calculate the Z-score is Z = (X - μ) / σ, where Z is the Z-score, X is the value, μ is the mean, and σ is the standard deviation.
In this case, we want to find the Z-score for the 90th percentile, which means that 90% of the population has a lower score. To find this Z-score, we can use a Z-table or a calculator.
Looking up the Z-score for the 90th percentile in a standard normal distribution table, we find it to be approximately 1.28. So, Z = 1.28.
Now, we can convert the Z-score back to the original scale using the formula X = (Z * σ) + μ, where X is the value, Z is the Z-score, σ is the standard deviation, and μ is the mean.
Let's substitute the values into the formula:
X = (1.28 * 12) + 88
X = 15.36 + 88
X = 103.36
Therefore, the score at the 90th percentile is approximately 103.36.
Hence, the correct option is c. 103.36.