Download The length of starfish in the Atlantic Ocean and Gulf of Mexico vary

The length of starfish in the Atlantic Ocean and Gulf
of Mexico vary according to the Normal distribution
with mean diameter of 24 cm and standard deviation
1.75 cm. Marine biologists classify starfish in to
multiple size categories: dwarf, medium and large.
Dwarf is defined as starfish with a diameter of 20.5
cm or less.
(a) What is the probability that a randomly selected
starfish is a dwarf, 20.5 cm or less?
(b) You choose 3 starfish at random; what is the
probability that the mean weight of the three
starfishes is considered dwarf, 20.5 cm or less?
(c) Explain why the probabilities in (a) and (b) are not
equal.
a. Normcdf (0, 20.5, 24, 1.75) = 0.02275
b. Normcdf (0, 20.5, 24, 1.75/ 3) =0.000266
c. A is the probability of finding 1 dwarf,
whereas B is the probability of finding 3
dwarfs. It is harder to find three extreme
values, than just 1.
The weight of the eggs produced by a certain breed
of hen is Normally distributed with mean 65 grams
(g) and standard deviation 5 g.
(a) Calculate the probability that a randomly
selected egg weighs between 62.5 g and 68.75 g.
Show your work.
(b) Think of cartons of such eggs as SRSs of size 12
from the population of all eggs. Calculate the
probability that the mean weight of the eggs in a
carton falls between 62.5 g and 68.75 g. Show your
work.
(c) Did you need to know that the population
distribution of egg weights was Normal in order to
complete parts (a) or (b)? Justify your answer.
a. Normcdf (62.5, 68.75, 65, 5) = 0.464835
b. Normcdf (62.5, 68.75, 65, 5/ 12) = 0.953677
c. Yes, in order to conduct Normal calculations
the distribution must be Normal or in the
case of sampling distributions the
distribution must have a sample size greater
than 30.
Suppose that 20% of a herd of cows is infected
with a particular disease. Assume to
independence condition is satisfied.
(a) What is the probability that the first diseased
cow is the 3rd cow tested?
(b) What is the probability that 4 or more cows
would need to be tested until a diseased cow
was found?
a. Geompdf (0.2, 3) = 0.128
b. Geomcdf (.2, 4, ∞)= 0.512
ACT scores for the 1,171,460 members of the
2004 high school graduating class who took the
test closely followed the Normal distribution with
mean 20.9 and standard deviation 4.8.
Choose two students independently and at
random from this group.
(a) What is the expected difference in their
scores?
(b) What is the standard deviation of the
difference in their scores?
(c) Find the probability that the difference in the
two students’ scores is greater than 6.
a. 0
b.
4.82 + 4.82 = 6.788
c. 1- Normcdf(-6, 6, 0, 6.788)= 1 - 0.581147 =
0.3705
Suppose an elderly person has two different
illnesses: cancer and Alzheimer's. Assume the
aliments are (unrealistically) independent. This
elderly person is enrolled in two different,
separate clinical trials to “cure” cancer and
Alzheimer’s. What is the probability the patient
is “cured” of both diseases? The probability that
the cancer cure fails is 87% and the probability
the Alzheimer’s cure fails is 93%.
0.13*0.07= 0.0091
The weights of oranges from a large orchard in
Florida are Normally distributed with a mean of
380 gm and a standard deviation of 28 gm.
(a) A single orange is selected at random from
this orchard. What is the probability that it
weighs more 400 gm?
(b) Three oranges are selected at random from
this orchard. What is the probability that their
mean weight is greater than 400 gm.?
(c) Explain why the probabilities in (a) and (b)
are not equal.
a. Normcdf(400, ∞, 380, 28) = 0.237525
b. Normcdf(400, ∞, 380, 28/ 3) =0.10801
c. A is the probability of finding 1 orange that is
big, whereas B is the probability of finding 3
oranges that are big. It is harder to find three
extreme values, than just 1 orange.