Answer :
Sure! Let's go through this step-by-step:
### 1. Describe the probability distribution of [tex]\( X \)[/tex] and state its parameters [tex]\(\mu\)[/tex] and [tex]\(\sigma\)[/tex].
The IQ scores are described as a normal distribution with a mean ([tex]\(\mu\)[/tex]) of 101.1 points and a standard deviation ([tex]\(\sigma\)[/tex]) of 11.7 points. So, the random variable [tex]\( X \)[/tex] representing the IQ scores follows a normal distribution, which we can denote as [tex]\( X \sim N(\mu = 101.1, \sigma = 11.7) \)[/tex].
### 2. Find the probability that the IQ score of a randomly selected person is:
#### a. Less than 98.1 points.
To find this probability, calculate the z-score for 98.1:
1. Calculate z-score:
[tex]\[ z = \frac{98.1 - 101.1}{11.7} = -0.2564 \][/tex]
2. Find the probability:
Using the standard normal distribution (z-table), the probability that [tex]\( X \)[/tex] is less than 98.1 is approximately 0.3988.
#### b. Between 78.5 and 83.2 points.
To find this probability, calculate the z-scores for both values:
1. Calculate z-scores:
For 78.5:
[tex]\[ z_{78.5} = \frac{78.5 - 101.1}{11.7} = -1.9368 \][/tex]
For 83.2:
[tex]\[ z_{83.2} = \frac{83.2 - 101.1}{11.7} = -1.5325 \][/tex]
2. Find the probabilities:
Find the cumulative probabilities for both z-scores and subtract the smaller one from the larger:
[tex]\[ P(78.5 < X < 83.2) = P(z \leq -1.5325) - P(z \leq -1.9368) \approx 0.0363 \][/tex]
#### c. More than 121.6 points.
To find this probability, calculate the z-score for 121.6:
1. Calculate z-score:
[tex]\[ z = \frac{121.6 - 101.1}{11.7} = 1.7564 \][/tex]
2. Find the probability:
The probability that [tex]\( X \)[/tex] is more than 121.6 is [tex]\( 1 - \text{cumulative probability for } z \)[/tex].
This is approximately 0.0399.
### 3. Find the 30th percentile for the IQ score.
The 30th percentile is the value below which 30% of the data falls.
1. Determine the value:
Use the standard normal distribution table or percentile function for the 30th percentile:
The 30th percentile for the normal distribution with a mean of 101.1 and a standard deviation of 11.7 is approximately 95.0 points.
So, summarizing, the answers are:
- Probability less than 98.1 points: [tex]\(0.3988\)[/tex]
- Probability between 78.5 and 83.2 points: [tex]\(0.0363\)[/tex]
- Probability more than 121.6 points: [tex]\(0.0399\)[/tex]
- 30th percentile: [tex]\(95.0\)[/tex] points
### 1. Describe the probability distribution of [tex]\( X \)[/tex] and state its parameters [tex]\(\mu\)[/tex] and [tex]\(\sigma\)[/tex].
The IQ scores are described as a normal distribution with a mean ([tex]\(\mu\)[/tex]) of 101.1 points and a standard deviation ([tex]\(\sigma\)[/tex]) of 11.7 points. So, the random variable [tex]\( X \)[/tex] representing the IQ scores follows a normal distribution, which we can denote as [tex]\( X \sim N(\mu = 101.1, \sigma = 11.7) \)[/tex].
### 2. Find the probability that the IQ score of a randomly selected person is:
#### a. Less than 98.1 points.
To find this probability, calculate the z-score for 98.1:
1. Calculate z-score:
[tex]\[ z = \frac{98.1 - 101.1}{11.7} = -0.2564 \][/tex]
2. Find the probability:
Using the standard normal distribution (z-table), the probability that [tex]\( X \)[/tex] is less than 98.1 is approximately 0.3988.
#### b. Between 78.5 and 83.2 points.
To find this probability, calculate the z-scores for both values:
1. Calculate z-scores:
For 78.5:
[tex]\[ z_{78.5} = \frac{78.5 - 101.1}{11.7} = -1.9368 \][/tex]
For 83.2:
[tex]\[ z_{83.2} = \frac{83.2 - 101.1}{11.7} = -1.5325 \][/tex]
2. Find the probabilities:
Find the cumulative probabilities for both z-scores and subtract the smaller one from the larger:
[tex]\[ P(78.5 < X < 83.2) = P(z \leq -1.5325) - P(z \leq -1.9368) \approx 0.0363 \][/tex]
#### c. More than 121.6 points.
To find this probability, calculate the z-score for 121.6:
1. Calculate z-score:
[tex]\[ z = \frac{121.6 - 101.1}{11.7} = 1.7564 \][/tex]
2. Find the probability:
The probability that [tex]\( X \)[/tex] is more than 121.6 is [tex]\( 1 - \text{cumulative probability for } z \)[/tex].
This is approximately 0.0399.
### 3. Find the 30th percentile for the IQ score.
The 30th percentile is the value below which 30% of the data falls.
1. Determine the value:
Use the standard normal distribution table or percentile function for the 30th percentile:
The 30th percentile for the normal distribution with a mean of 101.1 and a standard deviation of 11.7 is approximately 95.0 points.
So, summarizing, the answers are:
- Probability less than 98.1 points: [tex]\(0.3988\)[/tex]
- Probability between 78.5 and 83.2 points: [tex]\(0.0363\)[/tex]
- Probability more than 121.6 points: [tex]\(0.0399\)[/tex]
- 30th percentile: [tex]\(95.0\)[/tex] points
