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CAN SOMEONE HELP ME PLEASE?
well-known standard I.Q. test produces normally distributed results with a mean of 100 and a standard deviation of 15. What percent of the population have an I.Q. between 91 and 109?
a. 27.34%
b. 45.32%
PLEASE HELP
What is the domain of y=tan1 / 2 θ?
a. all real numbers
d. all real numbers except nπ, where n is an odd integer.
4 Answers
- llafferLv 72 years agoFavorite Answer
You do realize that there is a Details box that you can open up to add your details before you post it so you don't have to go back and add it as an update later, right?
This is a z-score problem, which requires the use of a z-score table (linked below).
It shows the probability of a random point of data being below a given point indicated by the z-score.
The z number is the number of standard deviations away from the mean, so a z-score of 0 is at the mean so 50% of the data is below it.
I use this equation to find z-score:
n = m + sz
Where n is the data point(s) in question (91 and 109)
m is the mean (100)
s is the standard deviation (15)
Solve for the unknowns:
n = m + sz
91 = 100 + 15z and 109 = 100 + 15z
-9 = 15z and 9 = 15z
-9/15 = z and 9/15 = z
-0.6 = z and 0.6 = z
When you look up a z-score in a table and gives you the probability of a value being below that number. So when you subtract the smaller from the larger if you look at the two graphs, then the overlap gets removed leaving you with the range that you are looking for (between the data points 91 and 109).
So now we have:
P(z = 0.6) - P(z = -0.6)
0.7257 - 0.2743
Now subtract and that's your answer as a probability. To turn it into a percent, multiply by 100:
0.4514
45.14%
The answer is closest to answer B. I'm not sure what rounding your book is expecting you to do, but this is the value that I get based on the table that I use.
Source(s): http://www.z-table.com/