Suppose scores on an IQ test are normally distributed. If the test has a mean of 100 and a standard deviation of 10, what is the probability that a person who takes the test will score between 90 and 110?
Check this answer that was posted yesterday.
http://www.jiskha.com/display.cgi?id=1254361096
Ms.Sue, you are amazing.
You're the amazing one, Reiny. You posted the answer with the link yesterday. :-)
Yes, but you found it so quickly.
I was still looking for it.
Ahh -- I'll let you in on a little secret. <g> I entered IQ test in the little Search box in the upper right corner of the Jiskha screen. Boom! Your post popped right up.
Thanks for the site. Is this the right formula:
P(-1 < z < 1) = 0.683
To find the probability that a person will score between 90 and 110 on an IQ test that is normally distributed with a mean of 100 and a standard deviation of 10, we can use the properties of the standard normal distribution.
1. Convert the raw scores to z-scores:
To convert a raw score to a z-score, we use the formula:
z = (x - μ) / σ
For a score of 90:
z1 = (90 - 100) / 10 = -1
For a score of 110:
z2 = (110 - 100) / 10 = 1
2. Find the probability associated with each z-score using a standard normal distribution table or a calculator. The table provides the area under the curve to the left of a given z-score.
Using the standard normal distribution table, the area to the left of a z-score of -1 is approximately 0.1587.
The area to the left of a z-score of 1 is also approximately 0.8413.
3. Calculate the probability of scoring between 90 and 110:
The probability of scoring between 90 and 110 can be calculated by subtracting the area to the left of a z-score of -1 from the area to the left of a z-score of 1.
P(90 ≤ X ≤ 110) = P(Z ≤ 1) - P(Z ≤ -1)
= 0.8413 - 0.1587
= 0.6826
Therefore, the probability that a person who takes the test will score between 90 and 110 is approximately 0.6826 or 68.26%.