Question 1084373: The frequency distribution for how long (in minutes) customers stayed to get service at xyz bank is shown below.
Time customer stayed
before service in xyz bank 1-10, 11-20, 21-30, 31-40, 41-50, 51-60
(in minutes)
Number of customers 11 , 8, 9, 7, 13, 2
Based on given information, find
a. The mean , mode and median minutes that customer stayed in the bank before service
b. The range, mean deviation, variance, standard deviation of the distribution
c. The probability that a customer will stay:
i. From 11 to 20 minutes
ii. No more than 40 minutes
iii. From 21 to 50 minutes
iv. For less than 50 minutes
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! If I use 5,15,25,35,45, and 55 as midpoints
5*11=55
15*8=120
25*9=225
35*7=245
45*13=585
55*2=110
sum is 1340
n=50
1340/50=26.8 mean.
median is half way between the 25th and 26th (half way between the halfway of an even number and the next one).
That is in the 20-30 interval or 25.
The mode is at 45.
range is between 5 and 55 or 50, with grouped data.
variance:
(-21.8^2)*11=5227.64
(-11.8^2)*8=1113.92
(-1.8^2)*9=29.16
(8.2^2)*7=470.68
(18.2^2)*13=4306.12
(28.2^2)*2=1590.48
sum is 12738
divide by 49259=259.96 variance
sqrt (259.96)=16.12 sd
=============================
11-20 minutes is 8/50=0.16
No more than 40 minutes is 35/70=0.7
From 21 to 50 minutes is 31/50=0.62
Less than 50 minutes is 48/50 or 0.96, because the interval middle is taken to be 45 for the interval containing 50.
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