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4 years ago

8

85,146 rounded to the nearest thousand

2 answers:

stiv31 [10]4 years ago

8 0

(Ignore this, my answer need to be 20 characters long)
the answer is 85,000

larisa [96]4 years ago

4 0

85,000 is your answer because The one in 146 is less then five

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3 years ago

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Answer:

Step-by-step explanation:

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4 years ago

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A meat inspector has randomly selected 30 packs of 95% lean beef. The sample resulted in a mean of 96.2% with a sample standard

kirza4 [7]

Answer:

a

The upper bound of the 99% prediction level is $98.2$

b

The 95% confidence interval is $9.7383 < \mu < 10.2617$

Step-by-step explanation:

Considering first question

From the question we are told that

The sample size is n = 30

The sample mean is $\= x = 96.2\%$

The standard deviation is $s = 0.8\%$

Generally the degree of freedom is mathematically represented as

$df = n - 1$

=> $df = 30 - 1$

=> $df = 29$

From the question we are told the confidence level is 99% , hence the level of significance is

$\alpha = (100 - 99 ) \%$

=> $\alpha = 0.01$

Generally from the t distribution table the critical value of at a degree of freedom of is

$t_{\alpha , 29} = 2.462$

Generally the 99% prediction level is mathematically represented as

$\= x \pm [(t_{\alpha , df }) * s * (\sqrt{1 + \frac{1}{ n} } )}]$

Generally the upper bound of the 99% prediction level is mathematically represented as

$\= x + [(t_{\alpha , df }) * s * (\sqrt{1 + \frac{1}{ n} } )}]$

=> $96.2 + (2.462 ) * 0.8 * (\sqrt{1 + \frac{1}{ 30} } )}]$

=> $98.2$

Considering second question

Generally the sample is mathematically represented as

$\= x = \frac{\sum x_i}{n}$

=> $\= x = \frac{ 9.8 + 10.2 + \cdots +9.6 }{7}$

=> $\= x = 10$

Generally the standard deviation is mathematically represented as

$\sigma = \sqrt{ \frac{ \sum ( x_ i - \= x)}{n-1} }$

=> $\sigma = \sqrt{ \frac{ ( 9.8 -10)^2 + ( 10.2 -10)^2 + \cdots + ( 9.6 -10)^2 }{7-1} }$

=> $\sigma = 0.283$

Generally the degree of freedom is mathematically represented as

$df = n- 1$

=> $df = 7- 1$

=> $df = 6$

From the question we are told the confidence level is 95% , hence the level of significance is

$\alpha = (100 - 95 ) \%$

=> $\alpha = 0.05$

Generally from the t distribution table the critical value of at a degree of freedom of is

$t_{\frac{\alpha }{2} , 6 } = 2.447$

Generally the margin of error is mathematically represented as

$E = t_{\frac{\alpha }{2} , 6 } * \frac{\sigma }{\sqrt{n} }$

=> $E =2.447* \frac{0.283 }{\sqrt{7} }$

=> $E =0.2617$

Generally 95% confidence interval is mathematically represented as

$\= x -E < \mu < \=x +E$

=> $10 -0.2617 < \mu < 10 + 0.2617$

=> $9.7383 < \mu < 10.2617$

7 0

3 years ago