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kherson [118]
3 years ago
10

This is important please help

Mathematics
1 answer:
Leya [2.2K]3 years ago
6 0

Answer: C

<u>Step-by-step explanation:</u>

∑ 8(\frac{7}{3})⁽ⁿ⁻¹⁾ from n = 1 to n = 5

n = 1 → 8(\frac{7}{3})⁽¹⁻¹⁾

        = 8                                         = \frac{648}{81}

n = 2 → 8(\frac{7}{3})⁽²⁻¹⁾

         = \frac{56}{3}                                         = \frac{1512}{81}    

n = 3 → 8(\frac{7}{3})⁽³⁻¹⁾

          = \frac{392}{9}                                     = \frac{3528}{81}

n = 4 → 8(\frac{7}{3})⁽⁴⁻¹⁾

          = \frac{2744}{27}                                 = \frac{8232}{81}

n = 5 → 8(\frac{7}{3})⁽⁵⁻¹⁾

          = \frac{19208}{81}                                = \frac{19208}{81}

                                                                            Sum = \frac{33128}{81}

                                                                                    = 408.99

     

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Let X the random variable the represent the scores for the test analyzed. We know that:

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The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Part a

For this case the mean and standard error for the sample mean would be given by:

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For this case we want this probability:

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P(\bar X \geq 285)=P(Z\geq \frac{285-276.1}{4.3}=2.070)

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P(Z\geq2.070)=1-P(Z

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