Calculate the standard score of the given x value, x = 59.4, where i = 66.2, s = 3.6. Round your answer to two decimal places. 13. %3D %3D Answer:

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### Statistics and Probability Exercise

#### Problem 11
**Question:**
Suppose that IQ scores have a bell-shaped distribution with a mean of 105 and a standard deviation of 15. Using the empirical rule, what percentage of IQ scores are greater than 120? Please do not round your answer.

**Solution:**
First, calculate the Z-score for 120 using the formula:
\[ Z = \frac{X - \mu}{\sigma} \]
where \( X = 120 \), \( \mu = 105 \), and \( \sigma = 15 \):
\[ Z = \frac{120 - 105}{15} = 1 \]

Next, we find the probability that \( Z \) is greater than 1. Using the empirical rule, the probability that a score is greater than 1 standard deviation above the mean is approximately 15.86553%.

**Answer:** 
\[ 15.86553 \% \]

#### Problem 12
**Question:**
Given the following data, find the weight that represents the 31st percentile.

**Data:** 
Weights of Newborn Babies (in pounds):

| 7.9 | 7.6 | 6.5 | 7.9 | 7.4 |
|----|----|----|----|----|
| 8.5 | 6.2 | 6.9 | 9.2 | 7.8 |
| 7.4 | 7.0 | 6.5 | 7.9 | 8.2 |

**Solution:**
Ordering the weights from smallest to largest and finding the 31st percentile involves calculation:
\[ 31\% \times 21 = 6.51 \approx 7^{th} \text{ value in the ordered list} \]

The ordered list is:
\[ 6.2, 6.5, 6.5, 6.9, 7.0, 7.4, 7.4, 7.6, 7.8, 7.9, 7.9, 7.9, 7.9, 8.2, 8.5, 9.2 \]

Therefore, the 7th value (which represents the 31st percentile) is:
\[ 6.9 \]

**Answer:**
\[ 6.9 \]
Transcribed Image Text:### Statistics and Probability Exercise #### Problem 11 **Question:** Suppose that IQ scores have a bell-shaped distribution with a mean of 105 and a standard deviation of 15. Using the empirical rule, what percentage of IQ scores are greater than 120? Please do not round your answer. **Solution:** First, calculate the Z-score for 120 using the formula: \[ Z = \frac{X - \mu}{\sigma} \] where \( X = 120 \), \( \mu = 105 \), and \( \sigma = 15 \): \[ Z = \frac{120 - 105}{15} = 1 \] Next, we find the probability that \( Z \) is greater than 1. Using the empirical rule, the probability that a score is greater than 1 standard deviation above the mean is approximately 15.86553%. **Answer:** \[ 15.86553 \% \] #### Problem 12 **Question:** Given the following data, find the weight that represents the 31st percentile. **Data:** Weights of Newborn Babies (in pounds): | 7.9 | 7.6 | 6.5 | 7.9 | 7.4 | |----|----|----|----|----| | 8.5 | 6.2 | 6.9 | 9.2 | 7.8 | | 7.4 | 7.0 | 6.5 | 7.9 | 8.2 | **Solution:** Ordering the weights from smallest to largest and finding the 31st percentile involves calculation: \[ 31\% \times 21 = 6.51 \approx 7^{th} \text{ value in the ordered list} \] The ordered list is: \[ 6.2, 6.5, 6.5, 6.9, 7.0, 7.4, 7.4, 7.6, 7.8, 7.9, 7.9, 7.9, 7.9, 8.2, 8.5, 9.2 \] Therefore, the 7th value (which represents the 31st percentile) is: \[ 6.9 \] **Answer:** \[ 6.9 \]
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