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%
Answer:
A
The test statistics is 
B
The Null and Alternative hypothesis are
and 
Step-by-step explanation:
From the question we are told that
The population mean is 
The sample size is 
The sample mean is 
The sample standard deviation is 
The level of significance is 
Given that the value which the manufacturer gave the automobile is 47.2 and it is believed that this is not correct, then
The Null Hypothesis is

The alternative Hypothesis is

The test statistics can be mathematically evaluated as

substituting values


We know that
if <span>4 electricians can complete a job in 8 hours
then
</span>100% of a job -------> 8 hours
x------------------> 1 hour
x=100/8----> x=12.5%
divide by 4 electricians
12.5%/4-----> 3.125%
that means each electrician can complete 3.125% of a job in 1 hour
therefore
3 electricians-----> 3*3.125%----> 9.375% in 1 hour
so
9.375%-----------> 1 hour
100%--------------> x
x=100/9.375----> x=10.67 hours
the answer is
10.67 hours
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