Answer:
0.6856
Step-by-step explanation:
![\text{The missing part of the question states that we should Note: that N(108,20) model to } \\ \\ \text{ } \text{approximate the distribution of weekly complaints).]}](https://tex.z-dn.net/?f=%5Ctext%7BThe%20missing%20part%20of%20the%20question%20states%20that%20we%20should%20Note%3A%20that%20%20N%28108%2C20%29%20model%20to%20%7D%20%5C%5C%20%5C%5C%20%20%5Ctext%7B%20%7D%20%5Ctext%7Bapproximate%20the%20distribution%20of%20weekly%20complaints%29.%5D%7D)
Now; assuming X = no of complaints received in a week
Required:
To find P(77 < X < 120)
Using a Gaussian Normal Distribution (
108, 
= 20)
Using Z scores:
As a result X = 77 for N(108,20) is approximately equal to to Z = -1.75 for N(0,1)
SO;
Here; X = 77 for a N(108,20) is same to Z = 0.6 for N(0,1)
Now, to determine:
P(-1.75 < Z < 0.6) = P(Z < 0.6) - P( Z < - 1.75)
From the standard normal Z-table:
P(-1.75 < Z < 0.6) = 0.7257 - 0.0401
P(-1.75 < Z < 0.6) = 0.6856
Hey there!
In order to answer this question, it's easiest to start with one of the money amounts and work your way to the other ones. Keep in mind that, no matter what, the amount of collective money won't exceed 1.50, meaning you can't have two 1 dollar bills, etc. When I wrote it out, I did it in this order:
1 = 1 dollar
0.5 = Half dollar
0.25 = Quarter
1 + 0.5 = 1.50
1 + 0.25 + 0.25 = 1.50
0.5 + 0.5 + 0.5 = 1.50
0.5 + 0.5 + 0.25 + 0.25 = 1.50
0.5 + (0.25*4) = 1.50
0.25*6 = 1.50
Since no other combinations add up to 1.50, your answer is 6.
Hope this helped you out! :-)
The correct answer is: 5.7109x10^6
I see no figure in your question
