IQ is normally distributed with a mean of 100 and a standard deviation of 15. Suppose one individual is randomly chosen. Let X = IQ of an individual. The middle 30% of IQs fall between what two values?
To find the values between which the middle 30% of IQs fall, we need to find the z-scores corresponding to the percentile values.
Step 1: Calculate the z-score corresponding to the lower percentile.
The lower percentile can be calculated as the average of the mean and the z-score representing the area to the left of the lower value. In this case, the lower percentile is (100 + z) where z is the z-score corresponding to the area to the left.
Since the middle 30% is symmetric, the lower tail would be (100-30)/2 = 35% or 0.35.
Using a standard normal distribution table or calculator, we can find the z-score corresponding to a cumulative probability of 0.35.
Looking up in the table, we find that the z-score is approximately -0.385.
Step 2: Calculate the z-score corresponding to the upper percentile.
The upper percentile can be calculated as the average of the mean and the z-score representing the area to the left of the upper value. In this case, the upper percentile is (100 + z) where z is the z-score corresponding to the area to the left.
Since the middle 30% is symmetric, the upper tail would be (100+30)/2 = 65% or 0.65.
Using a standard normal distribution table or calculator, we can find the z-score corresponding to a cumulative probability of 0.65.
Looking up in the table, we find that the z-score is approximately 0.385.
Step 3: Calculate the IQ values corresponding to the z-scores.
The IQ value corresponding to a z-score can be calculated using the formula: IQ = (z-score * standard deviation) + mean.
For the lower percentile: IQ = (-0.385 * 15) + 100 = 93.775 (approximated to 3 decimal places).
For the upper percentile: IQ = (0.385 * 15) + 100 = 106.775 (approximated to 3 decimal places).
Therefore, the middle 30% of IQs fall between the values of approximately 93.775 and 106.775.
To find the two values between which the middle 30% of IQs fall, we need to use the concept of z-scores and the standard normal distribution.
Step 1: Convert the IQ values to z-scores
To do this, we use the following formula:
z = (X - μ) / σ
where:
- X is the IQ value,
- μ is the mean IQ of 100, and
- σ is the standard deviation of 15.
Step 2: Find the z-scores corresponding to the 15th and 85th percentiles. The middle 30% corresponds to the area between the 15th and 85th percentiles.
Step 3: Convert the z-scores back to IQ values.
Let's calculate these values step by step.
Step 1: Convert IQ values to z-scores.
For the lower z-score (15th percentile):
z1 = (X1 - μ) / σ
= (X1 - 100) / 15
For the upper z-score (85th percentile):
z2 = (X2 - μ) / σ
= (X2 - 100) / 15
Step 2: Find the z-scores corresponding to the 15th and 85th percentiles.
From the standard normal distribution table or using statistical software, we find that the z-score corresponding to the 15th percentile is approximately -0.98, and the z-score corresponding to the 85th percentile is approximately 1.04.
So, z1 ≈ -0.98, and z2 ≈ 1.04.
Step 3: Convert the z-scores back to IQ values.
Using the formula for z-scores:
z = (X - μ) / σ
For the lower IQ value (X1):
-0.98 = (X1 - 100) / 15
Solving for X1:
X1 - 100 = -0.98 * 15
X1 - 100 = -14.7
X1 ≈ 85.3
For the upper IQ value (X2):
1.04 = (X2 - 100) / 15
Solving for X2:
X2 - 100 = 1.04 * 15
X2 - 100 = 15.6
X2 ≈ 115.6
Therefore, the middle 30% of IQs fall between the approximate values of 85.3 and 115.6.
You want Z scores that cut off .15 (15%) of the students either direction from the mean.
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion and put its Z score with the other values in the above equation. Solve for the score (both ±).