Use the table to answer each question. Note: Round z-scores to the nearest hundredth and then find the required A values using the table.
A psychologist finds that the intelligence quotients of a group of patients are normally distributed, with a mean of 102 and a standard deviation of 16. Find the percent of the patients with the following IQs?
(a) above 114
%
(b) between 90 and 120
To find the percent of patients with a specific IQ, we need to calculate the corresponding z-score and use the z-table to find the area under the normal distribution curve.
(a) To find the percent of patients with an IQ above 114:
1. Calculate the z-score using the formula: z = (X - μ) / σ, where X is the given IQ, μ is the mean, and σ is the standard deviation.
z = (114 - 102) / 16
z = 0.75 (rounded to the nearest hundredth)
2. Look up the z-score in the z-table to find the corresponding area. The z-score of 0.75 corresponds to an area of 0.7734.
3. Since we want to find the percentage above 114, subtract the area obtained in step 2 from 1, and then multiply by 100 to get the percentage:
Percentage = (1 - 0.7734) * 100
Percentage = 22.66% (rounded to the nearest hundredth)
Therefore, the percent of patients with an IQ above 114 is approximately 22.66%.
(b) To find the percent of patients with an IQ between 90 and 120:
1. Calculate the z-scores for the lower and upper bounds of 90 and 120, respectively:
z1 = (90 - 102) / 16
z1 = -0.75 (rounded to the nearest hundredth)
z2 = (120 - 102) / 16
z2 = 1.13 (rounded to the nearest hundredth)
2. Look up the z-scores in the z-table to find the corresponding areas.
The z-score of -0.75 corresponds to an area of 0.2266.
The z-score of 1.13 corresponds to an area of 0.8708.
3. To find the percentage between 90 and 120, subtract the smaller area obtained in step 2 from the larger area, and then multiply by 100:
Percentage = (0.8708 - 0.2266) * 100
Percentage = 64.42% (rounded to the nearest hundredth)
Therefore, the percent of patients with an IQ between 90 and 120 is approximately 64.42%.
To find the percent of patients with certain IQ scores, we need to use the z-score and the standard normal distribution table.
First, we need to calculate the z-scores for the given IQ scores. The formula for z-score is:
z = (X - μ) / σ
Where:
X is the value we want to find the percentage for (114 and 90 in this case),
μ is the mean of the distribution (102),
σ is the standard deviation (16).
(a) To find the percent of patients with IQs above 114:
Calculate the z-score using the formula:
z = (114 - 102) / 16 = 0.75
Next, we need to find the area under the normal distribution curve beyond z = 0.75. Look for the z-score of 0.75 in the table and find the corresponding area or percentile. The table usually provides the area to the left of the z-score.
The closest z-score in the table to 0.75 is 0.75. The corresponding area is 0.7734.
To find the area to the right of z = 0.75 (which is what we need), subtract the area we found from 1:
Area to the right of z = 0.75 = 1 - 0.7734 = 0.2266
Convert this to a percentage by multiplying by 100:
Percentage of patients with IQs above 114 = 0.2266 * 100 = 22.66%
Therefore, approximately 22.66% of the patients have IQs above 114.
(b) To find the percent of patients with IQs between 90 and 120:
Calculate the z-scores for 90 and 120 using the z-score formula:
For 90: z = (90 - 102) / 16 = -0.75
For 120: z = (120 - 102) / 16 = 1.125
The area to the left of z = -0.75 is 0.2266 and the area to the left of z = 1.125 is 0.8708.
To find the area between these two z-scores, subtract the smaller area from the larger area:
Area between z = -0.75 and z = 1.125 = 0.8708 - 0.2266 = 0.6442
Convert this to a percentage:
Percentage of patients with IQs between 90 and 120 = 0.6442 * 100 = 64.42%
Therefore, approximately 64.42% of the patients have IQs between 90 and 120.