Randy took a standardized test that is normally distributed. He scored a 105 on the test. The nationwide mean score on the test was 94 and the standard deviation was 6.2. What percent of the tested population scored higher than Randy on the test? Round to the nearest whole percent. O 2% O 4% O 96% O 94% O 11%
Inverse Normal Distribution
The method used for finding the corresponding z-critical value in a normal distribution using the known probability is said to be an inverse normal distribution. The inverse normal distribution is a continuous probability distribution with a family of two parameters.
Mean, Median, Mode
It is a descriptive summary of a data set. It can be defined by using some of the measures. The central tendencies do not provide information regarding individual data from the dataset. However, they give a summary of the data set. The central tendency or measure of central tendency is a central or typical value for a probability distribution.
Z-Scores
A z-score is a unit of measurement used in statistics to describe the position of a raw score in terms of its distance from the mean, measured with reference to standard deviation from the mean. Z-scores are useful in statistics because they allow comparison between two scores that belong to different normal distributions.
![**Problem:**
Randy took a standardized test that is normally distributed. He scored a 105 on the test. The nationwide mean score on the test was 94 and the standard deviation was 6.2.
**Question:**
What percent of the tested population scored higher than Randy on the test? Round to the nearest whole percent.
**Options:**
- 2%
- 4%
- 96%
- 94%
- 11%
**Solution Explanation:**
To determine what percentage of the population scored higher than Randy, we need to calculate the z-score and then use the standard normal distribution.
1. **Calculate the z-score:**
The z-score is calculated using the formula:
\[
z = \frac{(X - \mu)}{\sigma}
\]
Where:
- \( X \) is Randy's score (105)
- \( \mu \) is the mean score (94)
- \( \sigma \) is the standard deviation (6.2)
Substituting the values:
\[
z = \frac{(105 - 94)}{6.2} = \frac{11}{6.2} \approx 1.774
\]
2. **Use the z-score to find the corresponding percentile:**
Looking up the z-score of 1.774 in the standard normal distribution table, we find a cumulative probability of approximately 0.9616 (or 96.16%).
3. **Determine the percent scoring higher:**
The percent of the population scoring higher than Randy is:
\[
100\% - 96.16\% \approx 3.84\%
\]
Rounded to the nearest whole percent, the answer is:
**Answer:**
- 4%
The correct answer is 4%.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe1246267-81e7-461b-97b7-8e9aaefad2fd%2Fa3ab50d2-1f0c-434c-9df1-e0f2c8a55700%2Fdlbwarf_processed.jpeg&w=3840&q=75)
Trending now
This is a popular solution!
Step by step
Solved in 2 steps
