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DanielleElmas [232]
3 years ago
14

On average, a refrigerator door is opened 99 times each day. Len has two refrigerators in his house. Based on this average, abou

t how many times in a 8−week period are the refrigerator doors opened? The refrigerator doors are opened about times.
Mathematics
1 answer:
Firlakuza [10]3 years ago
8 0

Given :

On average, a refrigerator door is opened 99 times each day.

Len has two refrigerators in his house.

To Find :

How many times in a 8−week period are the refrigerator doors opened.

Solution :

Number of times door opened for 2 refrigerator , n = 99×2 = 198.

Number of days in 8 week , d = 8×7 = 56 days.

So, total number of times doors open in a period of 8-week is :

T=n\times d\\\\T=198\times 56\\\\T=11088\ times

Therefore, refrigerator doors are opened about 11088 times.

Hence, this is the required solution.

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Answer:

lower limit =0.261

upper limit =0.369

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

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Description in words of the parameter p

p represent the real population proportion of people that have experienced bullying

X= 63 people in the random sample that have experienced bullying

n=200 is the sample size required  

\hat p=\frac{63}{200}=0.315 represent the estimated proportion of people that have experienced bullying

z_{\alpha/2} represent the critical value for the margin of error  

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Numerical estimate for p

In order to estimate a proportion we use this formula:

\hat p =\frac{X}{n} where X represent the number of people with a characteristic and n the total sample size selected.

\hat p=\frac{63}{200}=0.315 represent the estimated proportion of people that they were planning to pursue a graduate degree

Confidence interval

The confidence interval for a proportion is given by this formula  

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}  

For the 90% confidence interval the value of \alpha=1-0.90=0.1 and \alpha/2=0.05, with that value we can find the quantile required for the interval in the normal standard distribution.  

z_{\alpha/2}=1.64  

And replacing into the confidence interval formula we got:  

0.315 - 1.64 \sqrt{\frac{0.315(1-0.315)}{200}}=0.261  

0.315 - 1.64 \sqrt{\frac{0.315(1-0.315)}{200}}=0.369  

And the 90% confidence interval would be given (0.261;0.369).  

We are confident at 90% that the true proportion of people that they were planning to pursue a graduate degree is between (0.261;0.369).  

lower limit =0.261

upper limit =0.369

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