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zalisa [80]
1 year ago
5

The weight of football players is normally distributed with a mean of 200 pounds and a standard deviation of 25 pounds.Refer to

Exhibit 6-3. What is the minimum weight of the middle 95% of the players?
Mathematics
1 answer:
Dafna11 [192]1 year ago
3 0

Given that

the weight of football players is distributed with a mean of 200 pounds and a standard deviation of 25 pounds.

And we need to find What is the minimum weight of the middle 95% of the players?

Explanation -

Using the Empirical Rule, 95% of the distribution will fall within 2 times of the standard deviation from the mean.

Two standard deviations = 2 x 25 pounds = 50 pounds

So the minimum weight = 200 pounds - 50 pounds = 150 pounds

Hence the final answer is 150 pounds.

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A company is approached by a new warehouse management systems vendor to adopt a new system. The vendor claims that the warehouse
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Answer:

a

     The null hypothesis  is  H_o  :  \mu  = 180

      The alternative hypothesis is  H_a :  \mu  <  180

b

The decision rule is  fail to reject the null hypothesis

The conclusion

There no sufficient evidence to support the vendor claim  that the warehouse management system reduces the average pick, pack, and ship time to below 3 minutes(180 seconds) per order through bin location and routing optimization

Step-by-step explanation:

From the question we are told that

     The population mean is  \mu  =  3 \  minutes =  180 \  seconds

     The sample  size is  n =  25

     The excel sheet is show on the first uploaded image

     The level of significance is  \alpha =  0.05

     The null hypothesis  is  H_o  :  \mu  = 180

      The alternative hypothesis is  H_a :  \mu  <  180

Generally the sample mean is mathematically represented as

        \= x  =  \frac{156 + 136 + \cdots  + 181}{25}

=>      \= x  = 173.2

Generally the standard deviation is mathematically represented as

      \sigma  =  \sqrt{\frac{\sum  (x_i - \= x )^2 }{n} }

=>    \sigma  =  \sqrt{\frac{  (156 - 173.2 )^2 + (136 - 173.2 )^2  + \cdots + (181- 173.2 )^2  }{25} }

=>    \sigma  =  24.261

Generally the test statistics is mathematically represented as

      t =  \frac{\= x  -  \mu }{\frac{\sigma}{\sqrt{n} } }

=>   t =  \frac{173.2 -  180 }{\frac{24.261}{\sqrt{25} } }

=>   z = -1.40    

Generally p-value is mathematically represented as

  p-value  =  P(Z <  z)

=>  p-value  =  P(Z < -1.40)

From the z-table  

   P(Z < -1.40) =  0.080757

=>   p-value  = 0.080757

From the obtained value we that  p-value >  \alpha

The decision rule is  fail to reject the null hypothesis

The conclusion

There no sufficient evidence to support the vendor claim  that the warehouse management system reduces the average pick, pack, and ship time to below 3 minutes(180 seconds) per order through bin location and routing optimization

   

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