SOLUTION: IQ scores in a certain population are normally distributed with a mean of 105 and a standard deviation of 14. (Give your answers correct to four decimal places.)
(a) Find the prob
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(a) Find the prob
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Question 924067: IQ scores in a certain population are normally distributed with a mean of 105 and a standard deviation of 14. (Give your answers correct to four decimal places.)
(a) Find the probability that a randomly selected person will have an IQ score between 104 and 110.
(b) Find the probability that a randomly selected person will have an IQ score above 99.
I don't have the slightest idea how to set it up, I don't know how to word it to find it online, and this book is not helping me.
Also, I don't know if it's just my computer/internet, but the topic box just shows a little with square with a black arrow. There are no words, so I have no clue what I just put this in. Answer by ewatrrr(24785) (Show Source):
You can put this solution on YOUR website! mean of 105 and a standard deviation of 14.
P( 104 < x< 110)
the Idea is to use z-values to find the Probability 0r Area Under the Standard Normal Curve
for ex:
For the normal distribution: Below: z = 0, z = ± 1, z= ±2 , z= ±3 are plotted.
Area under the standard normal curve to the left of the particular z is P(z)
Note: z = 0 (x value: the mean) 50% of the area under the curve is to the left and 50% to the right
P( 104 < x< 110) =
...
Using a TI calculator 0r similarly a Casio fx-115 ES plus
P( 104 < x< 110) = normalcdf( -1/14, 5/14) = normalcdf(-.0714, .3571) = .168
....
0r can be done: (Using tables from Your Book)
P( 104 < x< 110) = P(z < .3571) - P(z < -.0714) = .6395 - .4715 = .1680
....
P(x > 99) = P( z> -6/14) = P(z > -.4286) = 1 - P(z < -.4286) = 1 - .3341 = .6659
