The scores on a national achievement test are normally distributed with a mean of 500 and a standard deviation of 100. What percentage of those who took the test had a score greater than 630?
Answer (2)
Calculate the Z-score: Z = 100 630 − 500 = 1.3 .
Find the probability 1.3) = 1 - 0.9032 = 0.0968"> P ( Z > 1.3 ) = 1 − 0.9032 = 0.0968 .
Convert the probability to a percentage: 0.0968 × 100 = 9.68% .
The percentage of test takers who scored greater than 630 is 9.68% .
Explanation
Understand the problem and provided data The problem states that the scores on a national achievement test are normally distributed with a mean of 500 and a standard deviation of 100. We need to find the percentage of test takers who scored greater than 630.
Calculate the Z-score First, we need to calculate the Z-score, which represents how many standard deviations away from the mean a particular score is. The formula for the Z-score is: Z = σ x − μ where:
x is the score we are interested in (630 in this case),
μ is the mean of the distribution (500),
σ is the standard deviation of the distribution (100).
Compute Z-score value Plugging in the values, we get: Z = 100 630 − 500 = 100 130 = 1.3 So, a score of 630 is 1.3 standard deviations above the mean.
Find the corresponding probability Next, we need to find the probability of scoring greater than a Z-score of 1.3. We can use a standard Z-table or a calculator to find this probability. A Z-table gives the area to the left of the Z-score, so we need to subtract this value from 1 to find the area to the right (i.e., the probability of scoring greater than 630). From the Z-table, the area to the left of Z = 1.3 is approximately 0.9032. Therefore, the area to the right is: 1.3) = 1 - 0.9032 = 0.0968"> P ( Z > 1.3 ) = 1 − 0.9032 = 0.0968
Convert to percentage and conclude Finally, we convert this probability to a percentage by multiplying by 100: P erce n t a g e = 0.0968 × 100 = 9.68% Therefore, approximately 9.68% of those who took the test had a score greater than 630.
Examples
Understanding normal distributions and calculating probabilities is crucial in many real-world scenarios. For instance, in quality control, manufacturers use these concepts to determine the percentage of products that fall within acceptable quality ranges. Similarly, in finance, normal distributions help model stock returns and assess the likelihood of different investment outcomes. This problem demonstrates a fundamental application of statistics in analyzing and interpreting data.
Approximately 9.68% of the students who took the national achievement test scored greater than 630. This was calculated by determining the Z-score and finding the corresponding probability. The calculations showed that 1.3 standard deviations above the mean resulted in a score that only a small percentage of students exceeded.
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