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Answer: 49.85%
Step-by-step explanation:
Given : The physical plant at the main campus of a large state university recieves daily requests to replace florecent lightbulbs. The distribution of the number of daily requests is bell-shaped ( normal distribution ) and has a mean of 61 and a standard deviation of 9.
i.e. and
To find : The approximate percentage of lightbulb replacement requests numbering between 34 and 61.
i.e. The approximate percentage of lightbulb replacement requests numbering between 34 and .
i.e. i.e. The approximate percentage of lightbulb replacement requests numbering between and
. (1)
According to the 68-95-99.7 rule, about 99.7% of the population lies within 3 standard deviations from the mean.
i.e. about 49.85% of the population lies below 3 standard deviations from mean and 49.85% of the population lies above 3 standard deviations from mean.
i.e.,The approximate percentage of lightbulb replacement requests numbering between and
= 49.85%
⇒ The approximate percentage of lightbulb replacement requests numbering between 34 and 61.= 49.85%
The confidence interval comes out to be (0.685,0.735).
Calculating the Confidence Level and Other Terms
The confidence level can be calculated as follows,
z = %
z = 90 %
The margin error is given as, E= 0.025
The p value in this case is 0.71
Calculating the Confidence Interval
The mean of your estimate plus and minus the range of that estimate constitutes a confidence interval. Within a specific level of confidence, this is the range of values you anticipate your estimate to fall within if you repeat the test.
Confidence interval can be obtained by using the following formula,
(p-E, p+E) = (0.71-0.025, 0.71+0.025).
Therefore, the confidence interval is (0.685, 0.735).
Learn more about confidence interval here:
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