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. :

Answer:
monomial
Step-by-step explanation:
5x3=15. 15+6=21. The answer would be 3
Answer:
128
Step-by-step explanation:
Complete Question:
The complete question is shown on the first uploaded image
Answer:
The probability that the random you randomly select species that are greater than 200 kg is = 7/62
Step-by-step explanation:
Step One: Load the data set in to the R work space
data(mammals,package="MASS")
attach(mammals)
Step 2 : Obtain the list of the species that are greater than 200 and store it on y variable.
y <- subset(mammals,body>200)
Step Three : Obtain the total size
nrow(mammals)
Step Four : Obtain the sum of species greater than 200
sum(body > 200)
total size = 62
size with body > 200 = 7
hence
required probability = 7/62