assume adults have an IQ score that are normally distrubted with a mean of 105 and standard deviation of 20
A) find the percentage of adults that have an IQ between 94 and 116
B) find the percentage of adults IQ exceeding 120
C)Determine the 1st, 2nd, and 3rd quartiles for adults IQ
D) obtain the 95th percentile
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions related to the Z scores.
Reverse the process for the last two problems.
first Quartile = 25 percentile or lowest 25%
Second quartile = mean in normal distribution.
3rd quartile = lowest 75% percentile
0.3025
To solve these questions, we can use the Z-score formula and the Z-table.
The Z-score formula is given by:
Z = (X - μ) / σ
Where:
- X is the value we want to find the probability for,
- μ is the mean of the distribution (105 in this case),
- σ is the standard deviation of the distribution (20 in this case).
To find the probability using the Z-score, we can look up the Z-score in the Z-table.
A) To find the percentage of adults with IQ between 94 and 116, we need to find the probabilities for Z-scores corresponding to 94 and 116, and then subtract the lower probability from the higher probability.
1. Find the Z-score for 94:
Z₁ = (94 - 105) / 20 = -0.55
2. Find the Z-score for 116:
Z₂ = (116 - 105) / 20 = 0.55
Now, we can use the Z-table to find the probabilities corresponding to these Z-scores.
From the Z-table, the probability corresponding to Z₁ = -0.55 is 0.2910.
From the Z-table, the probability corresponding to Z₂ = 0.55 is 0.7089.
To find the probability between 94 and 116, we subtract the lower probability from the higher probability:
Probability = 0.7089 - 0.2910 = 0.4179 (approximately 41.79%).
Therefore, approximately 41.79% of adults have an IQ between 94 and 116.
B) To find the percentage of adults with an IQ exceeding 120, we calculate the probability of the Z-score being greater than the Z-score corresponding to 120.
1. Find the Z-score for 120:
Z = (120 - 105) / 20 = 0.75
From the Z-table, the probability corresponding to Z = 0.75 is 0.7734.
To find the probability of IQ exceeding 120, we subtract this probability from 1 (since we want the probability of the complement event):
Probability = 1 - 0.7734 = 0.2266 (approximately 22.66%).
Therefore, approximately 22.66% of adults have an IQ exceeding 120.
C) To find the quartiles of the IQ distribution, we can use the Z-scores.
The first quartile is the value below which 25% of the data lies. We can find its Z-score by looking up the Z-value for a probability of 0.25 in the Z-table:
Z₁₄ = -0.6745
To find the first quartile value (Q1), we can use the formula:
Q1 = μ + (Z₁₄ * σ)
Q1 = 105 + (-0.6745 * 20) = 92.51
Hence, the first quartile for the adults' IQ distribution is approximately 92.51.
The second quartile (median) is the value below which 50% of the data lies. It is also equal to the mean value. Therefore, the second quartile is 105.
The third quartile is the value below which 75% of the data lies. We can find its Z-score by looking up the Z-value for a probability of 0.75 in the Z-table:
Z₃₄ = 0.6745
To find the third quartile value (Q3), we can use the formula:
Q3 = μ + (Z₃₄ * σ)
Q3 = 105 + (0.6745 * 20) = 117.49
Hence, the third quartile for the adults' IQ distribution is approximately 117.49.
D) To find the 95th percentile, we need to find the Z-score corresponding to a probability of 0.95.
From the Z-table, the Z-value corresponding to a probability of 0.95 is approximately 1.645.
Using the formula for finding the 95th percentile:
X = μ + (Z * σ)
X = 105 + (1.645 * 20) = 137.9
Therefore, the 95th percentile for adults' IQ is approximately 137.9.