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?

Answer:

The probability of finding a sample mean less than 18 hours is 0.0082

Step-by-step explanation:

To find the probability of finding a sample mean less than 18 hours, we need to calculate the z-score of this sample mean 18. And the probability of finding a sample mean less than 18 hours is P(z<z(18)).

Z-score can be calculated as follows:

z(18)= where

• X is the sample mean (18 hours)

• M is the average hours dentists spend per week on fillings (20 hours)

• s is the standard deviation (10 hours)

• N is the sample size (144)

Putting the numbers, we get:

z(18)=

Using z- table we can find that P(z<z(18)) = 0.0082

Final answer:

Based on given mean & standard deviation, by using principles of Central Limit Theorem and Z-score calculation, the probability of finding a sample mean less than 18 hours is approximately 0.0082 or 0.82%.

Explanation:

This question is about the probability of a specific sample mean in statistics, based on provided mean and standard deviation values. It requires the principle of the Central Limit Theorem which states that means of samples taken from a population are normally distributed irrespective of the population's distribution.

To answer this question, we first need to calculate the standard error (SE), which is the standard deviation divided by the square root of the sample size (n). In this case, SE = 10/sqrt(144) = 10/12 = 0.83 (rounded to 2 decimal places).

Next, we calculate the Z score, which tells us how many standard deviations an element is from the mean. So, Z = (Sample Mean - Population Mean) / SE = (18 - 20) / 0.83 = -2.4 (rounded to one decimal place).

Using the Z score table (also known as a standard normal distribution table), we find that the probability of a Z value of -2.4 or less is approximately 0.0082. Thus, the probability of finding a sample mean less than 18 hours is 0.0082, or 0.82% when expressed as a percentage.

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