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3 years ago

Help Please I really don't understand how to do probability

Mathematics

1 answer:

ra1l [238]3 years ago

6 0

2/5+1/5=3/5
Or means add

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HELP ASAP (MATH)

alexdok [17]

PART - A : tent - 1 volume = 400 cubic ft

Tent -2 volume = 384 cubic ft.

PART - B : I recommend Joe to make the tent -1 with base 10x10 ft. as it has more volume

Step-by-step explanation:

PART - A:

tent 1 :

Base = 10 x 10ft

Height= 12 ft

Volume = (1/3) (10x 10) (12)

= (1/3) (100) (12)

= 400 cubic ft.

Tent 2:

Base = 12 x 12 ft

Height= 8 ft.

Volume = (1/3) (12 x 12) (8)

= 384 cubic ft.

PART - B:

I recommend Joe to make the tent -1 with base 10x10 ft. as it has more volume

8 0

3 years ago

A small lawnmower company produced 1,500 lawnmowers in 2008. In an effort to determine how maintenance-free these units were, th

ikadub [295]

Answer:

The 95% confidence interval for the average number of years until the first major repair is (3.1, 3.5).

Step-by-step explanation:

The (1 - α)% confidence interval for the average using the finite correction factor is:

$CI=\bar x\pm z_{\alpha/2}\cdot\frac{\sigma}{\sqrt{n}}\cdot\sqrt{\frac{N-n}{N-1}}$

The information provided is:

$N=1500\\n=183\\\sigma=1.47\\\bar x=3.3$

The critical value of z for 95% confidence level is,

z = 1.96

Compute the 95% confidence interval for the average number of years until the first major repair as follows:

$CI=\bar x\pm z_{\alpha/2}\cdot\frac{\sigma}{\sqrt{n}}\cdot\sqrt{\frac{N-n}{N-1}}$

$=3.3\pm 1.96\times\frac{1.47}{\sqrt{183}}\times\sqrt{\frac{1500-183}{1500-1}}\\\\=3.3\pm 0.19964\\\\=(3.10036, 3.49964)\\\\\approx (3.1, 3.5)$

Thus, the 95% confidence interval for the average number of years until the first major repair is (3.1, 3.5).

7 0

3 years ago

The CPA Practice Advisor reports that the mean preparation fee for 2017 federal income tax returns was $273. Use this price as t

skad [1K]

Answer:

a) 0.6212 = 62.12% probability that the mean price for a sample of 30 federal income tax returns is within $16 of the population mean.

b) 0.7416 = 74.16% probability that the mean price for a sample of 50 federal income tax returns is within $16 of the population mean.

c) 0.8804 = 88.04% probability that the mean price for a sample of 100 federal income tax returns is within $16 of the population mean.

d) None of them ensure, that one which comes closer is a sample size of 100 in option c), to guarantee, we need to keep increasing the sample size.

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 $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

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

The CPA Practice Advisor reports that the mean preparation fee for 2017 federal income tax returns was $273. Use this price as the population mean and assume the population standard deviation of preparation fees is $100.

This means that $\mu = 273, \sigma = 100$

A) What is the probability that the mean price for a sample of 30 federal income tax returns is within $16 of the population mean?

Sample of 30 means that $n = 30, s = \frac{100}{\sqrt{30}}$

The probability is the p-value of Z when X = 273 + 16 = 289 subtracted by the p-value of Z when X = 273 - 16 = 257. So

X = 289

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{289 - 273}{\frac{100}{\sqrt{30}}}$

$Z = 0.88$

$Z = 0.88$ has a p-value of 0.8106

X = 257

$Z = \frac{X - \mu}{s}$

$Z = \frac{257 - 273}{\frac{100}{\sqrt{30}}}$

$Z = -0.88$

$Z = -0.88$ has a p-value of 0.1894

0.8106 - 0.1894 = 0.6212

0.6212 = 62.12% probability that the mean price for a sample of 30 federal income tax returns is within $16 of the population mean.

B) What is the probability that the mean price for a sample of 50 federal income tax returns is within $16 of the population mean?

Sample of 30 means that $n = 50, s = \frac{100}{\sqrt{50}}$

X = 289

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{289 - 273}{\frac{100}{\sqrt{50}}}$

$Z = 1.13$

$Z = 1.13$ has a p-value of 0.8708

X = 257

$Z = \frac{X - \mu}{s}$

$Z = \frac{257 - 273}{\frac{100}{\sqrt{50}}}$

$Z = -1.13$

$Z = -1.13$ has a p-value of 0.1292

0.8708 - 0.1292 = 0.7416

0.7416 = 74.16% probability that the mean price for a sample of 50 federal income tax returns is within $16 of the population mean.

C) What is the probability that the mean price for a sample of 100 federal income tax returns is within $16 of the population mean?

Sample of 30 means that $n = 100, s = \frac{100}{\sqrt{100}}$

X = 289

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{289 - 273}{\frac{100}{\sqrt{100}}}$

$Z = 1.6$

$Z = 1.6$ has a p-value of 0.9452

X = 257

$Z = \frac{X - \mu}{s}$

$Z = \frac{257 - 273}{\frac{100}{\sqrt{100}}}$

$Z = -1.6$

$Z = -1.6$ has a p-value of 0.0648

0.9452 - 0.0648 =

0.8804 = 88.04% probability that the mean price for a sample of 100 federal income tax returns is within $16 of the population mean.

D) Which, if any of the sample sizes in part (a), (b), and (c) would you recommend to ensure at least a .95 probability that the same mean is withing $16 of the population mean?

None of them ensure, that one which comes closer is a sample size of 100 in option c), to guarantee, we need to keep increasing the sample size.

6 0

2 years ago

Which graph represents a function

tino4ka555 [31]

The last one represents a function ..

4 0

3 years ago

Read 2 more answers

Solve the equation for the variable. <img src="https://tex.z-dn.net/?f=%20-%205v%20%2B%209%20%3D%20%20-%20v%20%2B%2017" i

AlekseyPX

Answer:

v= - 2

Step-by-step explanation:

...............

8 0

3 years ago