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Jobisdone
12 days 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 [9.2K]12 days 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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Q13. A stone is dropped from a height of 5 km. The distance it falls through varies directly with the
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1 month ago
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What is the solution to the linear equation? 2.8y + 6 + 0.2y = 5y – 14 y = –10 y = –1 y = 1 y = 10
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2.8y + 6 + 0.2y = 5y – 14

Start by simplifying the left side:

3y + 6 = 5y - 14

Next, deduct 3y from both sides:

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2y = 20

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Write a difference quotient that best approximates the instantaneous rate of change of g at x=0
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26 days ago
Suppose your statistics professor reports test grades as​ z-scores, and you got a score of 1.57 on an exam. ​a) Write a sentence
tester [8842]

Answer:

a) In this case, we have a z-score of 1.57, characterized as:

z =\frac{x -\mu}{\sigma}

This signifies that our score is 1.57 standard deviations above the average of all test scores.

b) P(Z

Using the normal standard distribution or Excel, we computed:

P(Z

This represents 2.275% of the dataset.

Step-by-step explanation:

Previous concepts

Normal distribution denotes a "probability distribution that is symmetric around the mean, indicating that data close to the mean occurs more frequently than data further from it".

The Z-score serves as "a statistical measurement representing a value's relation to the average (mean) of a set, calculated in terms of standard deviations away from the mean".

Solution to the problem

Part a

For this instance, we hold a z-score of 1.57, which is defined as:

z =\frac{x -\mu}{\sigma}

This shows that our score is 1.57 deviations above the overall test score average.

Part b

A z-score of z=-2 indicates that your friend's score is 2 deviations below the other test scores.

Assuming a normal distribution, we can derive the percentage:

P(Z

Using the normal standard distribution or Excel, we discovered:

P(Z

This represents 2.275% of the data.

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