Let X be a binomial random variable with n = 100 and p = 0.2. Find approximations to these probabilities. (Round your answers to four decimal places.)

P(17 < X < 28)

I get the answer as 0.7059
but it is wrong.

I can't seem to figure this out... Help please

Nevermind! I figured out what I did wrong!

To find the probability P(17 < X < 28) for a binomial random variable, you can use the normal approximation to the binomial distribution.

To apply the normal approximation, you need to know the mean (μ) and standard deviation (σ) of the binomial distribution. For a binomial random variable, the mean is given by μ = n * p and the standard deviation is given by σ = sqrt(n * p * (1 - p)).

In this case, n = 100 and p = 0.2, so μ = 100 * 0.2 = 20 and σ = sqrt(100 * 0.2 * (1 - 0.2)) = sqrt(16) = 4.

Now, you can use the normal approximation to calculate the desired probability.

First, you need to standardize the values of 17 and 28 using the z-score formula: z = (x - μ) / σ.

For 17:
z1 = (17 - 20) / 4 = -0.75

For 28:
z2 = (28 - 20) / 4 = 2.00

Next, you need to find the probability associated with each z-score using the standard normal distribution table or calculator. Finding the probabilities associated with z1 and z2 will give you the bounds of the desired probability.

Using a standard normal distribution table, P(z < -0.75) = 0.2266 (rounded to four decimal places) and P(z < 2.00) = 0.9772 (rounded to four decimal places).

Therefore, to find P(17 < X < 28), you need to find the area between these two probabilities:

P(17 < X < 28) = P(z1 < z < z2) = P(z < z2) - P(z < z1) = 0.9772 - 0.2266 = 0.7506 (rounded to four decimal places).

So, the correct approximation to the probability P(17 < X < 28) is 0.7506, not 0.7059.