In Exercises 55–60, find the indicated probabilities and inter...

Question

In Exercises 55–60, find the indicated probabilities and interpret the results.

The mean annual salary for physical therapists in the United States is about $87,000. A random sample of 50 physical therapists is selected. What is the probability that the mean annual salary of the sample is (b) more than $85,000? Assume sigma = $10,500.

Accepted Answer

All right. Hello, everyone. So this question says, a company claims that the average delivery time for its packages is 48 hours, with a standard deviation of 4 hours. If a random sample of 32 deliveries is selected, what is the probability that the sample mean delivery time is more than 47 hours? And here we've got 4 different answer choices labeled A through D. So the first thing I want to point out here is that our sample size N is equal to 32. Which noticeably is greater than 30. According to the central limit theorem, if you recall. If a random sample has a size and that's greater than or equal to 30. And that sample is drawn from a population with a mean mu and a standard deviation sigma. The sampling distribution of sample means approximates a normal distribution. Which is the case here? So, therefore, the mean of sample means or mu of X bar is just equal to the population mean or mute. Which was given to us and is equal to 48. So now we can find the standard deviation of the sample meets or sigma of X bar. That's equal to sigma. Divided by the square root of N. So that's equal to or. Divided by the square root of 32. And this equals approximately 0.707. So for this question. What we're solving for is P of X bar being greater than 47. And before we can find this value, we have to know the Z score that corresponds to 47. So Z is equal to X bar Subtracted by me of X bar. Divided by sigma of X bar. This is equal to 47, subtracted by 48, and divided by 0.707. Ultimately, after rounding to three decimal places, this equals -1.414. So another way of expressing what we're solving for is P of Z being greater than. -1.414. And so to find What we're solving for our expression is equal to one subtracted by P of the less than -1.414. So, using the standard normal table, we would find the probability of Z less than -1.414. But what's noteworthy is that we don't exactly have this Z value in the table. Instead, our calculated Z value is between. 1.41. And -1.42. Which means that we need to interpolate. So the probability of. The less than. 1.42. Is 0.0778. Whereas for -1.41, that's 0.793. So by interpolation. Let me scroll down here to open up some space. By interpolation of. The less than 1.414. Is equal to. -1.414. Subtracted by -1.42. Divided by -1.41, subtracted by -1.42. Multiplied by 0.0793, subtracted by 0.0778. And then added to 0.0778. And this comes out to approximately 0.0787. So, substituting our interpolated value. P of Z greater than -1.414. is equal to 1 subtracted by 0.0787. And that equals approximately 0.9213, after rounding to four decimal places. So, in a random sample of 32 deliveries, the probability that the sample mean delivery time is greater than 47 hours. Is about 0.9213. This corresponds to option D, the multiple choice, making that your correct answer. And there you have it. So with that being said, thank you so very much for watching and I hope you found this helpful.

Key Concepts

Central Limit Theorem

Standard Error

Z-Score

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