Suppose that 65% of high school students eat breakfast before going to school. A random sample of 500 high school students were surveyed. Let X = the number of high school students in a random sample of size 500 who eat breakfast before going to school. 1. Explain why X can be modeled by a binomial distribution even though the sample was selected without replacement.

2. Use a binomial distribution to estimate the probability that 300 or fewer high school students in the sample eat breakfast before going to school.

3. Justify why X can be approximated by a Normal distribution.

4. Use a Normal distribution to estimate the probability that 300 or fewer high school students in the sample eat breakfast before going to school.

X can be modeled by a binomial distribution even though the sample was selected without replacement because the sample size is too large, and the probability that 300 or fewer high school students in the sample eat breakfast before going to school is 0.010809.

Given :

1) X can be modeled by a binomial distribution even though the sample was selected without replacement because the sample size is too large.

2) The formula for the binomial distribution is given by:

Now, substitute the known terms in the above formula.

Simplify the above expression.

3) Now, the value of np and n(1 - p) is given below:

np = 500 0.65 = 325

n(1 - p) = 500 (1 - 0.65) = 175

X can be approximated by a Normal distribution because the value of np and n(1 - p) is greater than or equal to 10.

4)

First, determine the standard deviation.

Now, the z-score can be calculated by using the formula:

z = -2.297

So, the probability that 300 or fewer high school students in the sample eat breakfast before going to school is given by:

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1.) very large sample size

2.) 0.11326

3.) np and n(1-p) should be ≥ 10

4.) 0.010809

Step-by-step explanation:

Given :

p = 0.65 ; n = 500

2.)

P(x ≤ 300)

P(x =x) = nCx * p^x * (1 - p)^(n - x)

Using a binomial probability calculator :

P(x ≤ 300) = 0.11326

To justify why if it can be approximated using a normal distributon :

np and n(1-p) should be ≥ 10

np = 500 * 0.65 = 325

n(1-p) = 500(1 - 0.65) = 175

4.) probability using normal distribution estimate:

Mean, m = np = 500 * 0.65 = 325

Standard deviation, s = sqrt(n*p*(1-p)) = sqrt(500*0.65*(1 - 0.65)) = sqrt(113.75) = 10.665

x ≤ 300

Correction:

x = x + 0.5 = 300 - 0.5 = 300.5

P(x ≤ 300.5)

Zscore = (x - m) / s

Zscore = (300.5 - 325) / 10.665

Zscore = - 24.5 / 10.665

Zscore = - 2.297

P(Z ≤ - 2.297) = 0.010809

1,645,493,059.35

step-by-step explanation:

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step-by-step explanation: