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. [tex]\mu = 6.3[/tex] minutes
[tex]\sigma=2.2[/tex] minutes
Let x be the long-distance call length.
a. The probability that a call lasts between 5 and 10 minutes will be :-
[tex]P(5<X<10)=P(\dfrac{5-6.3}{2.2}<\dfrac{X-\mu}{\sigma}>\dfrac{10-6.3}{2.2})\\\\=P(-0.59<Z<1.68)\ \ \ \ [z=\dfrac{X-\mu}{\sigma}]\\\\=P(z<1.68)-(1-P(z<0.59))\\\\=0.9535-(1-0.7224)\ \ \ \ [\text{by z-table}]\\\\=0.6759[/tex]
b. The probability that a call lasts more than 7 minutes. :
[tex]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<0.318)\\\\=1-0.6248\ \ \ \ [\text{by z-table}]\\\\=0.3752[/tex]
c. The probability that a call lasts more than 4 minutes. :
[tex]P(X<4)=P(\dfrac{X-\mu}{\sigma}<\dfrac{4-6.3}{2.2})\\\\=P(Z<-1.045)\ \ \ \ [z=\dfrac{X-\mu}{\sigma}]\\\\=1-P(Z<1.045)\\\\=1-0.8520 \ \ \ [\text{by z-table}]\\\\=0.1480[/tex]
