assume that adults have I.Q scores that are normally distributed with a mean of 100 and a standard deviation 15. find p2 which is the IQ score seperating bottom 2% from the top 98%
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 (.02) and its Z score. Insert Z score and other data in above equation and solve.
To find the IQ score that separates the bottom 2% from the top 98%, we need to find the corresponding Z-scores for these percentiles and then convert them back to IQ scores.
First, we need to calculate the Z-score using the formula:
Z = (X - μ) / σ
where:
X is the IQ score,
μ is the mean IQ score,
σ is the standard deviation of IQ scores,
Z is the Z-score.
For the bottom 2%:
P(Z < Z1) = 0.02
Using a Z-table or a calculator, you can find that Z1 is approximately -2.05.
To find X1, the corresponding IQ score, we rearrange the formula:
X1 = Z1 * σ + μ
X1 = -2.05 * 15 + 100
X1 = -30.75 + 100
X1 ≈ 69.25
So, an IQ score of approximately 69.25 separates the bottom 2% from the rest.
For the top 98%:
P(Z < Z2) = 0.98
Using a Z-table or a calculator, you can find that Z2 is approximately 2.05.
To find X2, the corresponding IQ score, we use the same formula:
X2 = Z2 * σ + μ
X2 = 2.05 * 15 + 100
X2 = 30.75 + 100
X2 ≈ 130.75
So, an IQ score of approximately 130.75 separates the top 98% from the rest.
Therefore, the IQ score separating the bottom 2% from the top 98% is approximately 69.25.
To find the IQ score that separates the bottom 2% from the top 98%, we can use the Z-score formula. The Z-score represents the number of standard deviations above or below the mean a particular data point is.
1. Convert the given percentile (2% or 0.02) to a Z-score using the standard normal distribution table or a Z-score calculator. For a one-sided distribution, we are interested in the left tail of the distribution, so we look for the Z-score that corresponds to the cumulative probability of 0.02.
The Z-score for a cumulative probability of 0.02 is approximately -2.05.
2. Use the Z-score formula to find the IQ score corresponding to the Z-score:
Z = (X - μ) / σ
Where:
Z is the Z-score
X is the IQ score we are looking for
μ is the mean IQ score (100 in this case)
σ is the standard deviation (15 in this case)
3. Rearrange the formula to solve for X:
X = Z * σ + μ
X = -2.05 * 15 + 100
X = -30.75 + 100
X ≈ 69.25
Therefore, the IQ score separating the bottom 2% from the top 98% is approximately 69.25. This means that 69.25 is the threshold below which only 2% of the population falls.