Question

 

A biologist estimates that the chance of germination for a type of bean seed is 0.8. A student was given 10 seeds. Let Xbe the number of seeds germinated from 10 seeds.

a) Assuming that the germination of seeds is independent, explain why the distribution of Xis binomial.

b) What are the values of nand p?

c) What is the mean and standard deviation of the number of seeds that germinate?

d) What is the probability that all seeds germinate?

e) What is the probability that only one seed does not germinate?

f) What is the probability that at most four seeds germinate?

g) Suppose that a group of students plant 200 seeds. What is the mean and standard deviation of the number that germinate?

h) What is the probability that at least 150 of the seeds germinate?

Answer 1

a)

It is given that germination of seeds is independent. For each seed, that is for each trial, the chance of germination for a type of bean seed is 0.8. There is only possible outcomes for each trial: either seed  germinate or not  germinate

So all conditions for binomial distribution has been satisfied so it is a binomial distribution.

b)

The values of n and p are

n = 10, p=0.80

c)

The mean is

The standard deviation is

d)

The probability that all seeds germinate is

e)

The probability that only one seed does not germinate, that is 9 seed germinate, is

f)

The probability that at most four seeds germinate is

g)

Here we have

n=200, p=0.8

The mean is

The standard deviation is

h)

Here we can normal approximation. The z-score for X = 150 -0.5 =149.5 is

The required probability is