For 8.4, only part c, for 8.10, only part b. 8.4 swer the following questions le

Question

For 8.4, only part c, for 8.10, only part b.
8.4 swer the following questions len being 19.62 mm An based ona r-8 population of bluegills with a known mean gth of 152.09 mm and a standard deviation r. a. What is the probability that any individual sampled at random from this population would have a length of 140 mm or shorter? b. What is the probability that a random sam- ple of 25 fish have a mean length of 140 mm or shorter? c. What is the probability that any individual sampled at random will have a length between 148 mm and 160 mm?

Explanation / Answer

Answer to the question is as follows:

8.4

Mean = 152.09

Stdev = 19.62

b. P(X<=140) = P(Z<= (140-152.09)/(19.62/sqrt(25))

= P(Z<=-3.08) = 0.001

So, there is just a .1% chance that a rs of 25 fish has mean length of 140 mm or shorter

c. P(148<X<160) = P((148-152.09)/19.62 <Z< (160-152.09)/19.62)

=P(-0.2085<Z<0.4032) = 0.6566-0.4174 = 0.2392

8.10. p(grey) = .53

n = 20

Mean = np = .53*20 = 1.06

Stdev = sqrt(npq) = sqrt(20*.53*.47) = 2.2320

P(X<=3) = P(X= 0,1,2,3)

= 20C0(.53^0)(.47^20) +..+ 20C3(.53^3)(.47^17)

=0.0005

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