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Answer :
Final answer:
The z-score for an IQ of 130 is 2, indicating that it is two standard deviations above the mean. This suggests that an IQ of 130 is higher than the majority of the population. The 95 percent confidence interval for an average IQ of 100 and a standard deviation of 15 is approximately from 70.6 to 129.4.
Explanation:
To find the z-score, we can use the formula: z = (X - μ) / σ, where X is the IQ score, μ is the population mean, and σ is the population standard deviation. For an IQ score of 130, with a population mean of 100 and a standard deviation of 15, the z-score can be calculated as follows:
z = (130 - 100) / 15 = 2
Therefore, the z-score is 2.
The number of standard deviations above the mean can be determined by the absolute value of the z-score. In this case, the z-score is 2, so there are 2 standard deviations above the mean.
With a z-score of 2 and considering that a z-score of 2 is beyond one standard deviation above the mean, it suggests that an IQ score of 130 is higher than the majority of the population.
The 95 percent confidence interval can be calculated using the formula: CI = X ± (1.96 * σ), where X is the sample mean. In this case, the average IQ of the population is 100 and the standard deviation is 15. Therefore, the 95 percent confidence interval is:
CI = 100 ± (1.96 * 15)
Calculating this, we get:
CI = 100 ± 29.4
So, the 95 percent confidence interval is approximately from 70.6 to 129.4.
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