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3 years ago

If i make 2230.80 per month, how much per hour for 40 hour work week

Mathematics

1 answer:

Dominik [7]3 years ago

7 0

2230.80 per month
=2230.80/4 = 557.70 per week
=557.70/40 = 13.9452 per hour

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In a sample of 60 electric motors, the average efficiency (in percent) was 85 and the standard deviation was 2. Section 05.01 Ex

jasenka [17]

Answer:

95% confidence interval for the mean efficiency is [84.483 , 85.517].

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We are given that in a sample of 60 electric motors, the average efficiency (in percent) was 85 and the standard deviation was 2.

So, the pivotal quantity for 95% confidence interval for the population mean efficiency is given by;

P.Q. = $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\mu$ = sample average efficiency = 85

$\sigma$ = sample standard deviation = 2

n = sample of motors = 60

$\mu$ = population mean efficiency

So, 95% confidence interval for the mean efficiency, $\mu$ is ;

P(-2.0009 < $t_5_9$ < 2.0009) = 0.95

P(-2.0009 < $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ < 2.0009 ) = 0.95

P( $-2.0009 \times {\frac{s}{\sqrt{n} }$ < ${\bar X - \mu}$ < $2.0009 \times {\frac{s}{\sqrt{n} }$ ) = 0.95

P( $\bar X -2.0009 \times {\frac{s}{\sqrt{n} }$ < $\mu$ < $\bar X +2.0009 \times {\frac{s}{\sqrt{n} }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X -2.0009 \times {\frac{s}{\sqrt{n} }$ , $\bar X +2.0009 \times {\frac{s}{\sqrt{n} }$ ]

= [ $85 -2.0009 \times {\frac{2}{\sqrt{60} }$ , $85 +2.0009 \times {\frac{2}{\sqrt{60} }$ ]

= [84.483 , 85.517]

Therefore, 95% confidence interval for the population mean efficiency is [84.483 , 85.517].

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