Question 1197409: In a large population of adults, the mean IQ is 122 with a standard deviation of 20. Suppose 50 adults are randomly selected for a market research campaign. (Round all answers to 4 decimal places, if needed.)
(a) The distribution of IQ is (approximately normal/exactly normal/ may or may not be normal/certainly skewed).
(b) The distribution of the sample mean IQ is (approximately normal/exactly normal/not normal/left skewed/ right skewed) with a mean of ___ and a standard deviation of___.
(c) The probability that the sample mean IQ is less than 121 is ___?
(d) The probability that the sample mean IQ is greater than 121 is ___?
(e) The probability that the sample mean IQ is between 121 and 125 is ___?
Answer by onyulee(41)   (Show Source):
You can put this solution on YOUR website! **a) Distribution of IQ**
* **May or may not be normal:**
* We are given that the population of adult IQ scores has a normal distribution.
* However, the distribution of individual IQ scores within the sample may or may not be perfectly normal due to random sampling variability.
**b) Distribution of the Sample Mean IQ**
* **Approximately Normal:**
* By the Central Limit Theorem, the distribution of the samplemean IQ will be approximately normally distributed, regardless of the underlying population distribution, as long as the sample size is sufficiently large (which is the case here with a sample size mean is equal to the population mean:
* Mean = 122
* **Standard Deviation of the Sample Mean (Standard Error):**
* Standard Deviation = σ / √n = 20 / √50 ≈ 2.8284
**c) Probability that the sample mean IQ is less than 121**
1. **Calculate the z-score:**
* z = (X̄ - μ) / (σ / √n) = (121 - 122) / 2.8284 = -0.3536
2. **Find the probability using a z-table:**
* P(X̄ < 121) = P(Z < -0.3536) = 0.3615
**d) Probability that the sample mean IQ is greater than 121**
* P(X̄ > 121) = 1 - P(X̄ < 121) = 1 - 0.3615 = 0.6385
**e) Probability that the sample mean IQ is between 121 and 125**
1. **Calculate the z-scores:**
* z1 = (121 - 122) / 2.8284 = -0.3536
* z2 = (125 - 122) / 2.8284 = 1.0607
2. **Find the probability using a z-table:**
* P(121 < X̄ < 125) = P(Z < 1.0607) - P(Z < -0.3536) = 0.8554 - 0 mean IQ is approximately normal with a mean of 122 and a standard deviation of 2.8284.
* P(X̄ < 121) = 0.3615
* P(X̄ > 121) = 0.6385
* P(121 < X̄ < 125) = 0.4939
**Note:** These calculations use the z-table (standard normal distribution table). You can also use statistical software like Excel or R to find these probabilities more precisely.
|
|
|
