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Elina [12.6K]
4 years ago
8

The physical plant at the main campus of a large state university recieves daily requests to replace florecent lightbulbs. The d

istribution of the number of daily requests is bell-shaped and has a mean of 57 and a standard deviation of 7. Using the empirical rule (as presented in the book), what is the approximate percentage of lightbulb replacement requests numbering between 50 and 57
Mathematics
1 answer:
krek1111 [17]4 years ago
8 0

Answer:

34%

Step-by-step explanation:

According to the empirical rule, 68% of all of the data in a normal distribution falls within one standard deviation of the mean.

Applying to this particular case, 68% of the requests fall within (50 - 7) and (50+ 7). The percentage of requests from 43 to 50 is equal to the percentage of requests from 50 to 57.

Therefore, the percentage of light bulb replacement requests numbering between 50 and 57 is 68/2 =  34%.

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A subway train is traveling at a rate of 22.4 m/s. Brakes are applied and it slows down at a constant rate of 3.5 m/s2 until it
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The diameters of aluminum alloy rods produced on an extrusion machine are known to have a standard deviation of 0.0001 in. A ran
Murljashka [212]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The null hypothesis is rejected this means that \mu \ne  0.5025

The 95% confidence interval is 0.5045608 <  \mu  < 0.5046392

Step-by-step explanation:

From the question we are told that

The standard deviation is s= 0.0001

The sample size is n = 25

The sample mean is \= x =  0.5046 \ in

The population mean is \mu  =  0.5025 \ in

The null hypothesis is H_o  :  \mu =  0.5025

The alternative hypothesis is H_a  :  \mu \ne  0.5025

The test statistics is mathematically represented as

t =  \frac{0.5046 -  0.5025}{ \frac{0.0001}{\sqrt{25} } }

t =  105

So the p-value from the z-table is mathematically represented as

p-value  =  2 *  P( z >  105)

p-value  = 0.000

seeing that

p-value <  \alpha we reject the null hypothesis

The critical value of

\frac{\alpha }{2} obtained from the normal distribution table is

Z_{\frac{\alpha }{2} } =  1.96

The margin of error is mathematically represented as

E = Z_{\frac{\alpha }{2} }*\frac{s}{\sqrt{n} }

=> E = 1.96 *\frac{0.0001}{\sqrt{25} }

=> E =3.92 *10^{-5}

The 95% confidence level is mathematically represented as

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

=> 0.5046  -  3.92 *10^{-5}  <  \mu  < 0.5046  +  3.92 *10^{-5}

=> 0.5045608 <  \mu  < 0.5046392

8 0
3 years ago
Question 5 Unsaved
kicyunya [14]

Its  {3, 3, 6, 7, 9}

because the difference in values of 6  and the other numbers are the same as difference between 5 and the other numbers in { 2, 2, 5, 6, 8}

8 0
3 years ago
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