Question

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 1

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)=$\frac{X-M}{\frac{s}{\sqrt{N} } }$ 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)=$\frac{18-20}{\frac{10}{\sqrt{144} } }=-2.4$

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