Question

Mirchi an FM stereo station in india, finds that the mean length of time a person is tuned to the station is 17.0 minutes with a standard deviation of 3 minutes. what is the probability that a sample of size 64 will result into a sample mean that is:

a) Between 16 to 16.5 minutes

b) Less than 16 but more than 18.125 minutes.

Answer 1

Solution

Mean Length = 17.0 minutes

Standard deviation = 3 minutes

Sample Size = 64

Find sample mean

a.) Between 16 to 16.5 minutes

$P(16<\bar{x}<16.5)=P(\frac{16-17}{\frac{3}{\sqrt{64}}}<z<\frac{16.5-17}{\frac{3}{\sqrt{64}}})\\ =P(-\frac{8}{3}<z<-\frac{4}{3})\\ =P(z<-\frac{4}{3})-P(z<-\frac{8}{3})\\ =P(z<1.33)-P(z<2.67)\\ \therefore \implies0.09176-0.00379 = 0.08797$

0.08797 is the mean length of time a person is tuned to the station between 16 to 16.5 minutes

b.) Less than 16 but more than 18.125 minutes

$P(\bar{x}<16) \cap P(\bar{x}>18.125)=P(z<\frac{16-17}{\frac{3}{\sqrt{64}}})+P(z>\frac{18.125-17}{\frac{3}{\sqrt{64}}})\\ =P(z<-2.67)+P(z>3)\\ \therefore 0.00379+0.09987=0.10366$

Hence the mean length of the time a person is tuned to the station in less than 16 minutes but more than 18.125 is 0.10366