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

Round 50.057 to the nearest hundredth.

icang [17]

50.057 = 50.06 <span>to the nearest hundredth.</span>

8 0

4 years ago

Read 2 more answers

What is the opposite of the opposite of -7

Alex Ar [27]

The answer is -7

no to sure though

5 0

3 years ago

Read 2 more answers