On the package for a certain brand of spinach seeds there is a guarantee that, if the printed instructions are followed, of planted seeds will germinate. A random sample of seeds is chosen. If these seeds are planted according to the instructions, find the probability that of them germinate.

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.

P(4<X<5) = 0.624

Step-by-step explanation: The probability of a seed germinate is a Binomial Distribution, 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.

answer for #1 is c

answer for #2 is d

can i get brainliest

b

step-by-step explanation: