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Jobisdone
2 months ago
14

A national dental association conducted a survey to find the average (mean) amount of time dentists spend on dental fillings per

week. Based on a simple random sample, they surveyed 144 dentists. The statistics showed that dentists spent an average of 20 hours per week on fillings with a standard deviation of 10 hours. What is the probability of finding a sample mean less than 18 hours?
Mathematics
1 answer:
lawyer [12.5K]2 months ago
4 0
The likelihood of observing a sample mean that is less than 18 hours is 0.0082. \nTo evaluate this probability, we calculate the z-score for a sample mean of 18. Accordingly, the probability of getting a sample mean below 18 hours becomes P(z<z(18)). \nThe z-score is calculated as follows: \nz(18) = [(X - M) / s] where: \n- X is the sample mean (18 hours) \n- M is the average hours dentists devote weekly to fillings (20 hours) \n- s is the standard deviation (10 hours) \n- N is the sample size (144) \nSubstituting the numbers leads to: \nz(18) = [(18 - 20) / (10/sqrt(144))]. Using the z-table, we find that P(z<z(18)) is 0.0082.
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