Question

Consider the probability that greater than 94 out of 153 people will not get the flu this winter. Assume the probability that a given person will not get the flu this winter is 65%. Approximate the probability using the normal distribution. Round your answer to four decimal places.

Answer 1

To find the probability of more than 94 out of 153 people not getting the flu, we use the normal distribution approximation. The mean and standard deviation are calculated based on the given probabilities. We then use the normal distribution table or a calculator to find the probability.

Step-by-step explanation:

To find the probability that greater than 94 out of 153 people will not get the flu, we can use the normal distribution approximation. First, we need to calculate the mean and standard deviation of the number of people who will not get the flu.

The mean, μ, can be calculated by multiplying the total number of people by the probability that a given person will not get the flu, which is 65%. So, μ = 153 * 0.65 = 99.45.

The standard deviation, σ, is equal to the square root of the number of people multiplied by the probability of not getting the flu multiplied by the probability of getting the flu. So, σ = sqrt(153 * 0.65 * 0.35) = 4.86.

Using these values, we can calculate the probability using the normal distribution table or a calculator. We want to find the probability of getting more than 94 people not getting the flu, which is equivalent to finding the area to the right of 94 when using the normal distribution curve. Round the answer to four decimal places.

Answer 2

0.8212 = 82.12% probability that greater than 94 out of 153 people will not get the flu this winter.

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

`E(X) = np`

The standard deviation of the binomial distribution is:

`√(V(X)) = √(np(1-p))`

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that
`\mu = E(X)`,
`\sigma = √(V(X))`.

In this problem, we have that:

`p = 0.65, n = 153`. So

`\mu = E(X) = 153*0.65 = 99.45`

`\sigma = √(V(X)) = √(np(1-p)) = √(153*0.65*0.35) = 5.9`

Consider the probability that greater than 94 out of 153 people will not get the flu this winter

This probability is 1 subtracted by the pvalue of Z when X = 94. So

`Z = (X - \mu)/(\sigma)`

`Z = (94 - 99.45)/(5.9)`

`Z = -0.92`

`Z = -0.92` has a pvalue of 0.1788

1 - 0.1788 = 0.8212

0.8212 = 82.12% probability that greater than 94 out of 153 people will not get the flu this winter.