The IQs of army volunteers in a given year are normally distributed wi...
Solution:
Given that:
Mean IQ (μ) = 110
Standard Deviation (σ) = 10
Percentage of recruits to be selected for advanced training = 20%
To find:
What is the lowest IQ score acceptable for the advanced training?
Step 1: Find the Z-score for the given percentage
Z-score is the number of standard deviations an observation or data point is from the mean. It is given by the formula:
Z = (X - μ) / σ
where X is the raw score, μ is the mean and σ is the standard deviation.
We need to find the Z-score for the 80th percentile (i.e., the top 20% of the scores). From the standard normal distribution table, we know that the Z-score for the 80th percentile is 0.84.
Step 2: Find the IQ score corresponding to the Z-score
We can use the same formula as above to find the IQ score corresponding to the Z-score.
Z = (X - μ) / σ
0.84 = (X - 110) / 10
X - 110 = 8.4
X = 118.4
Therefore, the lowest IQ score acceptable for advanced training is 118.4.
Explanation:
In a normal distribution, the majority of the scores are clustered around the mean, and as we move away from the mean, the number of scores decreases. The standard deviation gives us an idea of how spread out the scores are from the mean.
In this problem, we are given the mean and standard deviation of the IQ scores of army volunteers. We are also told that the army wants to give advanced training to the top 20% of the recruits.
To find the IQ score corresponding to the 80th percentile, we need to find the Z-score for the 80th percentile from the standard normal distribution table. Once we have the Z-score, we can use the formula Z = (X - μ) / σ to find the IQ score corresponding to the Z-score.
In this case, the Z-score is 0.84, which means that the IQ score corresponding to the 80th percentile is 118.4. Therefore, any recruit with an IQ score of 118.4 or higher will be eligible for advanced training.
The IQs of army volunteers in a given year are normally distributed wi...
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