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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.
Answer:
(19/99) = 0.192
Step-by-step explanation:
Numbers 1 to 99, that is 99 numbers (obtained through the equation for the nth term of an AP)
L = a + (n-1) d
L = nth term = 99
a = first term = 1
n = number of terms = ?
d = common difference = 1
99 = 1 + (n-1)1
n = 99
Sample space = 99
The numbers that include at least, a 1 are
1, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 21, 31, 41, 51, 61, 71, 81, 91
19 numbers.
Probability of randomly selecting a card where the number contains at least one digit 1 from 1 to 99 = (19/99) = 0.192
Hope this Helps!!!
Answer:
Below are the three ways.
Hope I helped :)
