Question

assume that all samples are simple random samples and that all populations are normally distribute.

A female Tuna fish have a mean of 316 lbs and a standard deviation of 51lbs

a. find probability (4 dp) that 1 randomly-selected adult female tuna weighs between 260lbs and 300 lbs. draw a bell-shaped curve and shade the area corresponding to the probability.

b. find the probability (4dp) that a sample of 7 female tuna has a mean weight between 260lbs and 300 lbs. draw a bell-shaped curve and shade the area corresponding to the probability.

c. find the cut off (1 dp) separating the heaviest 12% of female tuna from the rest. (please round in a way that no more than the heaviest 12% of adult female tuna are at or above the above cut off in weight) keep answer in pounds and draw the bell-shaped curve and shade in the area corresponding to the referred to percentage in the exercise)

Answer 1

1)

for normal distribution z score =(X-μ)/σx
here mean=       μ=	316
std deviation   =σ=	51.000

probability =P(260<X<300)=P((260-316)/51)<Z<(300-316)/51)=P(-1.1<Z<-0.31)=0.3769-0.1361=0.2408

b)

sample size       =n=	7
std error=σx̅=σ/√n=	19.2762

probability =P(260<X<300)=P((260-316)/19.276)<Z<(300-316)/19.276)=P(-2.91<Z<-0.83)=0.2033-0.0018=0.2014

c)

for top 12 % or 88th percentile critical value of z=		1.1750
therefore corresponding value=mean+z*std deviation=			375.9