Question

the white hot pepper is a traditional jazz band.the length in minutes of each piece of music played by the band may be modelled by a normal distribution with mean 5 and standard deviation 1.5 and may be assumed to be independent of the length of all other pieces.

find the probability that a particular piece will last between 3.5 and 7.25 minutes

find the probability that the average length of the next 6pieces to be played by the band will be less than 4 minutes.

there is an interval of exactly one minute btw the band finishing one piece and starting the next. the band starts to play its last 6pieces at 10:31pm. using your answer above state whether you think the band will still be playing at 11:00pm. justify your answer

Answer 1

a. Let $X=$ the length in minutes of each piece of music played by the band: $X\sim N(\mu , \sigma^2).$

Given $\mu=5, \sigma=1.5$

$$P(3.5<X<7.25)=P(X<7.25)-P(X\leq3.5)$$

$$=P(Z<\dfrac{7.25-5}{1.5})-P(Z\leq\dfrac{3.5-5}{1.5})$$

$$=P(Z<1.5)-P(Z\leq-1)$$

$$\approx0.9331928-0.1586553$$

$$\approx0.7745$$

b. Let $\bar{X}=$ the average length i: $\bar{X}\sim N(\mu , \sigma^2/n).$

Given $\mu=5, \sigma=1.5, n=6$

$$P(\bar{X}<4)=P(Z<\dfrac{4-5}{1.5/\sqrt{6}})$$

$$\approx P(Z<-1.633)\approx0.0512$$

c. Let $Y=$ the length of 6 pieces with intervals: $Y=6X+5$

$Y\sim N(\mu_1, \sigma_1^2)$

$$\mu_1=6(5)+5=35$$

$$\sigma_1^2=(6)^2(\sigma)^2=81$$

$$P(Y>29)=1-P(Y\leq29)$$

$$=1-P(Z\leq \dfrac{29-35}{9})\approx1-P(Z\leq -0.6667)$$

$$\approx0.7475$$

Since $0.7475>0.7,$ I think the band will still be playing at 11:00pm.