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mart [117]
4 years ago
7

The teacher marked Silvano's problem wrong on his test.

Mathematics
1 answer:
mote1985 [20]4 years ago
7 0
I͟ h͟a͟v͟e͟ n͟o͟ i͟d͟e͟a͟ w͟h͟a͟t͟ h͟e͟ d͟i͟d͟ w͟r͟o͟n͟g͟ b͟e͟c͟a͟u͟s͟e͟ I͟ d͟o͟n͟'t͟ k͟n͟o͟w͟ t͟h͟e͟ o͟r͟i͟g͟i͟n͟a͟l͟ p͟r͟o͟b͟l͟e͟m͟.
B͟u͟t͟, f͟r͟o͟m͟ w͟h͟a͟t͟ I͟ c͟a͟n͟ s͟e͟e͟,

(4^5)^4= 4^5•4

=>4^20

W͟o͟u͟l͟d͟ b͟e͟ t͟h͟e͟ c͟o͟r͟r͟e͟c͟t͟ a͟n͟s͟w͟e͟r͟.

N͟o͟w͟, I͟ c͟a͟n͟ s͟e͟e͟ S͟i͟l͟v͟a͟n͟o͟ d͟o͟i͟n͟g͟ t͟h͟i͟s,

(4^5)^4=4^5+4

=>4^9

B͟u͟t͟, w͟h͟e͟n͟ a͟ p͟r͟o͟b͟l͟e͟m͟ i͟s͟ p͟u͟t͟ i͟n͟ p͟a͟r͟e͟n͟t͟h͟e͟s͟e͟s͟,
t͟h͟e͟n͟ r͟a͟i͟s͟e͟d͟ t͟o͟ t͟h͟e͟͟ p͟o͟w͟e͟r͟ o͟f͟ a͟ n͟u͟m͟b͟e͟r͟,
M͟U͟L͟T͟I͟P͟L͟Y͟ t͟h͟e͟ e͟x͟p͟o͟n͟e͟n͟t͟s͟ t͟o͟g͟e͟t͟h͟e͟r͟.


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A manufacturer makes chocolate squares that have a target weight of 8 g. Quality control engineers sample 30 chocolate squares t
posledela

Using the <em>normal distribution and the central limit theorem</em>, it is found that the power of the test is of 0.9992 = 99.92%.

<h3>Normal Probability Distribution</h3>

In a normal distribution with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

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

  • It 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, which is the percentile of X.
  • By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation s = \frac{\sigma}{\sqrt{n}}.

In this problem:

  • The mean is \mu = 8.5.
  • The standard deviation is \sigma = 0.87.
  • A sample of 30 is taken, hence n = 30, s = \frac{0.87}{\sqrt{30}} = 0.1588.

The power of the test is given by the probability of a sample mean above 8, which is <u>1 subtracted by the p-value of Z when X = 8</u>, so:

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

By the Central Limit Theorem:

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

Z = \frac{8 - 8.5}{0.1588}

Z = -3.15

Z = -3.15 has a p-value of 0.0008.

1 - 0.0008 = 0.9992.

The power of the test is of 0.9992 = 99.92%.

To learn more about the <em>normal distribution and the central limit theorem</em>, you can check brainly.com/question/24663213

5 0
3 years ago
What is sum of two fifths and one forth?
Alecsey [184]
2/5+1/4 = 8+5/20 = 13/20 ...answer !!!
3 0
4 years ago
I NEED HELP ASAP!!!!!!!!!!!!!!!!
Snowcat [4.5K]

Answer:

1/6

Step-by-step explanation:

If you divide the fractions to decimals 1/6 equals .16 which is closest to 0

7 0
3 years ago
Read 2 more answers
A normally distributed random variable with mean 4.5 and standard deviation 7.6 is sampled to get two independent values, X1 and
mr Goodwill [35]

Answer:

Bias for the estimator = -0.56

Mean Square Error for the estimator = 6.6311

Step-by-step explanation:

Given - A normally distributed random variable with mean 4.5 and standard deviation 7.6 is sampled to get two independent values, X1 and X2. The mean is estimated using the formula (3X1 + 4X2)/8.

To find - Determine the bias and the mean squared error for this estimator of the mean.

Proof -

Let us denote

X be a random variable such that X ~ N(mean = 4.5, SD = 7.6)

Now,

An estimate of mean, μ is suggested as

\mu = \frac{3X_{1} + 4X_{2}  }{8}

Now

Bias for the estimator = E(μ bar) - μ

                                    = E( \frac{3X_{1} + 4X_{2}  }{8}) - 4.5

                                    = \frac{3E(X_{1}) + 4E(X_{2})}{8} - 4.5

                                    = \frac{3(4.5) + 4(4.5)}{8} - 4.5

                                    = \frac{13.5 + 18}{8} - 4.5

                                    = \frac{31.5}{8} - 4.5

                                    = 3.9375 - 4.5

                                    = - 0.5625 ≈ -0.56

∴ we get

Bias for the estimator = -0.56

Now,

Mean Square Error for the estimator = E[(μ bar - μ)²]

                                                             = Var(μ bar) + [Bias(μ bar, μ)]²

                                                             = Var( \frac{3X_{1} + 4X_{2}  }{8}) + 0.3136

                                                             = \frac{1}{64} Var( {3X_{1} + 4X_{2}  }) + 0.3136

                                                             = \frac{1}{64} ( [{3Var(X_{1}) + 4Var(X_{2})]  }) + 0.3136

                                                             = \frac{1}{64} [{3(57.76) + 4(57.76)}]  } + 0.3136

                                                             = \frac{1}{64} [7(57.76)}]  } + 0.3136

                                                             = \frac{1}{64} [404.32]  } + 0.3136

                                                             = 6.3175 + 0.3136

                                                              = 6.6311

∴ we get

Mean Square Error for the estimator = 6.6311

6 0
3 years ago
Hey <br> Please someone help me!!!!!<br> Asap
storchak [24]
(0,4) and (1,0)
Hope this helps
8 0
3 years ago
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