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 :-
![P(5](https://tex.z-dn.net/?f=P%285%3CX%3C10%29%3DP%28%5Cdfrac%7B5-6.3%7D%7B2.2%7D%3C%5Cdfrac%7BX-%5Cmu%7D%7B%5Csigma%7D%3E%5Cdfrac%7B10-6.3%7D%7B2.2%7D%29%5C%5C%5C%5C%3DP%28-0.59%3CZ%3C1.68%29%5C%20%5C%20%5C%20%5C%20%5Bz%3D%5Cdfrac%7BX-%5Cmu%7D%7B%5Csigma%7D%5D%5C%5C%5C%5C%3DP%28z%3C1.68%29-%281-P%28z%3C0.59%29%29%5C%5C%5C%5C%3D0.9535-%281-0.7224%29%5C%20%5C%20%5C%20%5C%20%5B%5Ctext%7Bby%20z-table%7D%5D%5C%5C%5C%5C%3D0.6759)
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. :
![P(X](https://tex.z-dn.net/?f=P%28X%3C4%29%3DP%28%5Cdfrac%7BX-%5Cmu%7D%7B%5Csigma%7D%3C%5Cdfrac%7B4-6.3%7D%7B2.2%7D%29%5C%5C%5C%5C%3DP%28Z%3C-1.045%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%3C1.045%29%5C%5C%5C%5C%3D1-0.8520%20%5C%20%5C%20%5C%20%5B%5Ctext%7Bby%20z-table%7D%5D%5C%5C%5C%5C%3D0.1480)
Answer:
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Step-by-step explanation:
The correct answer is C. F(x)=2 * (0.7)^x
Answer:
2400 slabs. If this helps, please give brainliest!