It is believed that 4% of children have a gene that may be linked to juvenile diabetes. Researchers at a firm would like to test new monitoring equipment for diabetes. Hoping to have 21 children with the gene for their study, the researchers test 733 newborns for the presence of the gene linked to diabetes. What is the probability that they find enough subjects for their study?

the probability that they find enough subjects for their study is 0.9515

Step-by-step explanation:

From the given information:

Let X be the random variable that follows a normal distribution.

Therefore:

X Binomial(n=733,p=0.04)

The probability of P(X ≥ 21) lies in the region between 20.5 and 21.5 by considering a discrete contribution of a continuous normal distribution. Eventually, X > 20.5

∴

P(X >20.5)= P(Z > z)

Using  standard normal z formula:

z = -1.66

P(X >20.5)= P(Z > -1.66)

From the standard z tables ;

P(X >20.5)= 1 - 0.0485

P(X >20.5)= 0.9515

question 1 is 8, question 2 is 9, question 3 is 24

step-by-step explanation:

now, the numbers they give you at the top are the key. from there, you just have to plug and chug! let's try 1, 2, and 3!

key: a=3, b=5, c=6

1. a+5

remember, a=3, so we plug that in to get 3+5=8

2. 15-c

c=6, so we plug that in to get 15-6=9

3. 4b

b=6, so we plug it in to get 4*6=24

time to try the rest on your own! if you put the answers below in a comment, i'll check them for you!

step-by-step explanation: