Danielle is keeping track of the growth of a basil plant. The table below shows the measurements she made on different days. Day
1 answer:
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Answer:
a) Binomial distribution with parameters p=0.85 q=0.15 n=6
b) 62.29%
c) 2.38%
d) See explanation below
Step-by-step explanation:
a)
We could model this situation with a binomial distribution
where P(6;k) is the probability of finding exactly k people out of 6 with Rh positive, p is the probability of finding one person with Rh positive and q=(1-p) the probability of finding a person with no Rh.
So
b)
The probability that Y is less than 6 is
P(Y=0)+P(Y=1)+...+P(Y=5)
Let's compute each of these terms
and adding up these values we have that the probability that Y is less than 6 is
c)
In this case is a binomial distribution with n=200 instead of 6.
p and q remain the same.
The mean of this sample would be 85% of 200 = 170.
In a binomial distribution, the standard deviation is
In this case
<em>Let's approximate the distribution with a normal distribution with mean 170 and standard deviation 5.05</em>
So, the approximate probability that there are fewer than 160 persons with Rh positive blood in a sample of 200 would be the area under the normal curve to the left of 160
(see picture attached)
We can compute that area with a computer and find it is
0.0238 or 2.38%
d)<em> In order to approximate a binomial distribution with a normal distribution we need a large sample like the one taken in c).</em>
In general, we can do this if the sample of size n the following inequalities hold:
in our case np = 200*0.85 = 170 and nq = 200*0.15 = 30
