The question is incomplete. Here is the complete question.
On the package for a certain brand of spinach seeds there is a guarantee that, if the printed instructions are followed, 63% of planted seeds will germinate. A random sample of 9 seeds is chosen. If these seeds are planted according to the instructions, find the probability that 4 or 5 of them germinate. Do not round your intermiediate computations, and round your answer to three decimal places.
Answer: P(4<X<5) = 0.624
Step-by-step explanation: The probability of a seed germinate is a <u>Binomial</u> <u>Distribution</u>, i.e., a discrete probability distribution of the number of successes in a sequence of n independents experiments.
This distribution can be approximated to normal distribution by determining the values of mean and standard deviation population:
where
n is the sample quantity
p is proportion of successes
For the spinach seeds:
Mean is
5.85
Standard deviation is
1.431
Now, use
to convert into a standard normal distribution.
The probability we want is between 2 values: P(4<X<5).
Therefore, we have to convert those two values:
For X = 4:
z = -1.29
For X = 5:
z = -0.59
Using z-table:
P(X>4) = 1 - P(z< -1.29) = 0.9015
P(X<5) = P(z< -0.59) = 0.2776
The probability will be
P(4<X<5) = P(X>4) - P(X<5)
P(4<X<5) = 0.9015 - 0.2776
P(4<X<5) = 0.624
The probability of 4 or 5 seeds germinate is 0.624.
