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. $\mu = 6.3$ minutes

$\sigma=2.2$ minutes

Let x be the long-distance call length.

a. The probability that a call lasts between 5 and 10 minutes will be :-

$P(5$

b. The probability that a call lasts more than 7 minutes. :

$P(X>7)=P(\dfrac{X-\mu}{\sigma}>\dfrac{7-6.3}{2.2})\\\\=P(Z>0.318)\ \ \ \ [z=\dfrac{X-\mu}{\sigma}]\\\\=1-P(Z$

c. The probability that a call lasts more than 4 minutes. :

$P(X$

8 0

3 years ago

There are 6 red marbles, 8 blue marbles, and 11 green marbles in a bag. What is the probability that you

Kay [80]

Answer:

14/25

Step-by-step explanation:

p(red or blue)=6/(6+8+11) +8/(6+8+11)=6/25+8/25=14/25

7 0

3 years ago

Can someone help me with this question?

Lina20 [59]

Answer:1/3

Step-by-step explanation:

gradient equals

$\frac{Y2-Y1}{X2-X1}$

= $\frac{5-3}{10-4}$

= $\frac{2}{6}$

= $\frac{1}{3}$

6 0

2 years ago