Answer: a. 0.6759 b. 0.3752 c. 0.1480
Step-by-step explanation:
Given : The long-distance calls made by the employees of a company are normally distributed with a mean of 6.3 minutes and a standard deviation of 2.2 minutes
i.e. 
minutes
minutes
Let x be the long-distance call length.
a. The probability that a call lasts between 5 and 10 minutes will be :-
b. The probability that a call lasts more than 7 minutes. :
![P(X>7)=P(\dfrac{X-\mu}{\sigma}>\dfrac{7-6.3}{2.2})\\\\=P(Z>0.318)\ \ \ \ [z=\dfrac{X-\mu}{\sigma}]\\\\=1-P(Z](https://tex.z-dn.net/?f=P%28X%3E7%29%3DP%28%5Cdfrac%7BX-%5Cmu%7D%7B%5Csigma%7D%3E%5Cdfrac%7B7-6.3%7D%7B2.2%7D%29%5C%5C%5C%5C%3DP%28Z%3E0.318%29%5C%20%5C%20%5C%20%5C%20%5Bz%3D%5Cdfrac%7BX-%5Cmu%7D%7B%5Csigma%7D%5D%5C%5C%5C%5C%3D1-P%28Z%3C0.318%29%5C%5C%5C%5C%3D1-0.6248%5C%20%5C%20%5C%20%5C%20%5B%5Ctext%7Bby%20z-table%7D%5D%5C%5C%5C%5C%3D0.3752)
c. The probability that a call lasts more than 4 minutes. :
Y = -x2 + 5x + 36 <span>→ y = -(x2 -5x -36)
</span><span>→ y = -(x2 - 9x +4x - 36)
</span><span>→ y = -[x(x-9) + 4(x - 9)]
</span><span>→ y = -(x - 9)(x + 4)
Your answer would be </span>y=-(x-9)(x+4).
I’m sorry if I’m wrong but I do think your answer is X=22y^-1
X is always dependant to Y. so that should answer the 1st one. the 2nd looks to be around 17-18. 17.5 maybe. and the 3rd is yes cause he does not go to the pool so it isnt any cost.
The only possible outcome to this would be any number greater or less than 361 in the 300’s
