Answer:
Mailing preparation takes 38.29 min max time to prepare the mails.
Step-by-step explanation:
Given:
Mean:35 min
standard deviation:2 min
and 95% confidence interval.
To Find:
In normal distribution mailing preparation time taken less than.
i.eP(t<x)=?
Solution:
Here t -time and x -required time
mean time 35 min
5 % will not have true mean value . with 95 % confidence.
Question is asked as ,preparation takes less than time means what is max time that preparation will take to prepare mails.
No mail take more time than that time .
by Z-score or by confidence interval is
Z=(X-mean)/standard deviation.
Z=1.96 at 95 % confidence interval.
1.96=(X-35)/2
3.92=(x-35)
X=38.29 min
or
Confidence interval =35±Z*standard deviation
=35±1.96*2
=35±3.92
=38.29 or 31.71 min
But we require the max time i.e 38.29 min
And by observation we can also conclude the max time from options as 38.29 min.
Most likely number 3
we can guess the average of the two since we don't have the time to actually add all of them and divide and number 3 describes it the most
L=w+4m
A=w*l
A=w(w+4)=45
w^2+4w-45=0
w^2+9w-5w-45=0
w(w+9)-5(w+9)=0
(w+9)(w-5)=0
⇒ w+9=0 or w-5=0
⇒w-5=0
w=5
l=5+4
l=9
To find an unknown number take the end number and add it to the number you already have.
2 + 2 = 4
x = 4
