Suppose a batch of metal shafts produced in a manufacturing company have a standard deviation of 2 and a mean diameter of 200 inches.

If 83 shafts are sampled at random from the batch, what is the probability that the mean diameter of the sample shafts would differ from the population mean by less than 0.2 inches? Round your answer to four decimal places.

Answers

Answer 1

Answer:

0.6372 = 63.72% probability that the mean diameter of the sample shafts would differ from the population mean by less than 0.2 inches.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]

Standard deviation of 2 and a mean diameter of 200 inches.

This means that [tex]\sigma = 2, \mu = 200[/tex]

83 shafts

This means that [tex]n = 83, s = \frac{2}{\sqrt{83}}[/tex]

What is the probability that the mean diameter of the sample shafts would differ from the population mean by less than 0.2 inches?

Mean between 200 - 0.2 = 199.8 inches and 200 + 0.2 = 200.2 inches, which is the p-value of Z when X = 200.2 subtracted by the p-value of Z when X = 199.8.

X = 200.2

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{200.2 - 200}{\frac{2}{\sqrt{83}}}[/tex]

[tex]Z = 0.91[/tex]

[tex]Z = 0.91[/tex] has a p-value of 0.8186

X = 199.8

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{199.8 - 200}{\frac{2}{\sqrt{83}}}[/tex]

[tex]Z = -0.91[/tex]

[tex]Z = -0.91[/tex] has a p-value of 0.1814

0.8186 - 0.1814 = 0.6372

0.6372 = 63.72% probability that the mean diameter of the sample shafts would differ from the population mean by less than 0.2 inches.


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Step-by-step explanation:

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The circumference of the base of the cone is 8.5 inches. What is the volume of the cone in term of pi? Round to the nearest hundredth

Answers

Answer:  V = 270.94/pi

========================================================

Work Shown:

C = 2*pi*r  ......... circumference of the circular base

r = C/(2pi)

r = (8.5)/(2pi)

r = 4.25/pi

-------------

V = pi*r^2*h ...... volume of the cone

V = pi*(4.25/pi)^2*15

V = pi*(18.0625/pi^2)*15

V = (18.0625*15)/pi

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If we round that decimal number up top to the nearest hundredth, then we end up with V = 270.94/pi

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Answers

Part (a)

Refer to figure 1 in the attached image below. The a,b,c,d,e are placeholders for numbers we'll fill in later (that turns into the table for figure 2). Hopefully you agree about the placement of the numbers. If there's any confusion, then please let me know.

Along the bottom row, we can see that 39+e = 65 which solves to e = 26 after you subtract 39 from both sides. This tells us that there are 26 non-physics majors (ie their major is anything but physics).

Then along the "non-physics" column, notice how b+6 = e, which is the same as b+6 = 26 after replacing 'e' with 26. Solving that equation leads to b = 20. So we have 20 people who are seniors and non-physics majors.

Now move to the top row. We can see that a+b = 30, which is the same as a+20 = 30 because we plug in b = 20. That solves to a = 10. So we have 10 seniors who are physics majors.

----------------

Now along the first column, we have a+c = 39, aka 10+c = 39, which solves to c = 29. There are 29 non-seniors who are physics majors.

The last variable to find is variable d.

Along the second row, we can say,

c+6 = d

29+6 = d

d = 35

Or we could say that 30+d = 65 along the last column which also solves to d = 35. This tells us we have 35 non-seniors in the class.

----------------

To summarize so far, we have

a = 10, b = 20, c = 29, d = 35, e = 26

These values will replace the letters as shown in figure 2.

Figure 2 will be used to answer both parts (a) and (b).

----------------

After that (more than) slight detour of filling in the table, we can finally tackle the question at hand. We're asked to find the probability of selecting someone who is a senior and a physics major. We have a = 10 people who fit the description out of 65 total

Divide the two values and reduce the fraction as much as possible

10/65 = (2*5)/(13*5) = 2/13

Answer:   2/13

==========================================================

Part (b)

There's a lot going on with part (a). Luckily there aren't many steps here because pretty much all the work has been done already. We'll refer to figure 2 below.

We're told "given the student is a physics major", which means we focus solely on the "physics major" column only. We ignore everyone else because we know 100% that whoever we selected majored in physics.

We have a = 10 seniors who are also physics majors out of 39 physics majors total.

That constructs the fraction 10/39

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Step-by-step explanation:

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Step-by-step explanation & answer:

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Answers

Answer:

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Answers

Answer:

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Step-by-step explanation:

The population of a rabbit colony triples each month.

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9514 1404 393

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