Final answer:
For a given population mean IQ of 100 and standard deviation of 15, the correct sample mean IQ with only a 0.01 probability of being as high or higher by random sampling with a sample size of 25 is 107.84.Option B is corect.
Step-by-step explanation:
The original question presented requires the use of statistical methods to determine a sample mean IQ that has only a 0.01 probability of being as high or higher by chance. The mean (µ) of the IQ distribution is given as 100, with a standard deviation (σ) of 15. When looking for a certain percentile in a normal distribution, we employ the Z-score formula. Considering the sample size (n) of 25, we would use the standard normal distribution to find the corresponding Z-score that leaves 0.01 of the distribution to the right of it. Once the Z-score is found using standard statistical tables or software, the sample mean can be calculated using the formula for the sampling distribution of the sample mean: μ + (Z*σ/√n). Given the choices presented, B) 107.84 is the correct option for the sample mean IQ that has a 0.01 probability of being as high or higher by chance.
Regarding the distribution to use for the hypothesis test on the claim that the mean IQ score on the Stanford-Binet IQ test is more than 100, with a known standard deviation of all scores being 15 points and a sample size of 15, the correct distribution would be B) Student's tdistribution. This is because the sample size is less than 30, which is typically the threshold for using the normal distribution in such hypothesis testing when the population standard deviation is known.