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Statistics HW Help?
The basal diameter of a sea anemone is an indicator of its age, and in a certain
population of anemones, the distribution of basal diameters is approximately normal
with a mean of 5.3 cm and a standard deviation of 1.8 cm. Suppose you randomly
select five anemones from this population.
a.
What is the probability that all five anemones have a basal diameter more than 5.5
cm?
(2pt)
b.
What is the probability that the mean basal diameter of the five anemones is more
than 5.5 cm?
Comparing the probabilities you calculated in parts (a) and (b), which is larger?
Even if we changed the values the mean and standard deviation of this population, briefly explain why this must will always be the case.
1 Answer
- AlanLv 74 years ago
a.
What is the probability that all five anemones have a basal diameter more than 5.5
cm?
(2pt)
Here you find the probability of one anemone is greater than 5.5
then you use binomial probability to
find the probability of all 5 being greater.
for one anemones
P( X> 5.5) = 1- P(x< 5.5)
P(X< 5.5 ) = P ( Z < ( x - mean) /standard deviation ) = P (z< (5.5 -5.3) / 1.8) =
P(X< 5.5) = P( z < 0.2/1.8) = P(z< (1/9) ) = P(z< 0.111111111 )
https://www.stat.tamu.edu/~lzhou/stat302/standardn...
Per this table,
P(z< 0.11) = .54380
P(z< 0.12 ) = .54776
=== Interpolating (Some professor and classes tell just the closest value. so if you not supposed
to interpolating use 0.54380 )
=+0.54380 + ( 0.54776-0.54380)*(1/9) = 0.54424
P(x< 5.8) = 0.54424
P(x> 5.8) = 1 - 0.54424 = 0.45576
Now remember binomial probability
P( k successes in n trials ) = (n k ) *(p)^k*(1-p)^(n-k)
where (n k ) = (n!) / ( (n-k)! * k!) =
but we are looking at (n = 5 , k = 5, p = 0.45576)
(5 5) = 5! / ( 5!) *(0!) = 1
===== answer part (a)
P( 5 successes ) = p^k = (0.45576)^5 = 0.019664415
b.
What is the probability that the mean basal diameter of the five anemones is more
than 5.5 cm?
Since the original distribution is normal and you know the population standard deviation .
You can use the normal distribution for the sample mean here.
mean of the sample mean = population mean = 5.3
standard deviation of the sample mean = population standard deviation / sqrt (N)
standard deviation of the sample mean = 1.8/sqrt(5) = 0.804984472
so now use standard equation with mean = 5.3 and standard deviation = 0.804984472
P(x> 5.5) = 1 - P(x<5.5)
P(x< 5.50) = p(Z < (5.5 -5.3) / 0.804984472 ) = p(Z< 0.248451997)
P(z< 0.24 ) = .59483
P(z< 0.25 ) = .59871
so if you interpolate
=0.59483 + ( 0.248451997 - 0.24 ) * ( 0.59871 -0.59483)/ 0.01 =
= 0.598109375
rounding the 5 digits of the table
= 0.59811
P(x> 5.8) = 1-0.59811 = 0.401890625
==== answer (B)
=0.40189
Extra Question
(b) is larger obviously
why is b always larger
for (a) , you are taking your answer for one trial and then raising it to 5th power .
Whenever you square any number which is between 0 and 1 , the result is a smaller number.
Probability always only go from 0 to 1 ( 0 and 1 included). However , the 0 and 1 cases are
really abnormal conditions.
This is even worse when you take a number between 0 and 1 and take it to the 5th power.
Each times you multiple the probability against itself, the results get smaller.
So the answer for (a) will always be smaller.
If the question was P(x< 5.5) , it would be even simpler, but since it is P(x> 5.5) it makes it a little
more complicated.