Question

7% of Americans have an O negative blood type. A SRS of 400 Americans are surveyed. Using normal approximation to the binomial, what is the approximate probability that at least 34 of them have the o negative blood type?

Please show work!

Answer 1

Let denote the random variable denoting the number of Americans who have a blood group of O negative.

It is given that a sample of 400 Americans has been considered and approximately 7% of the Americans have O negative blood type. Therefore, the random variable follows the Binomial distribution with parameters and .

The mean, , and the variance, , of the random variable X is calculated below:

And,

In the given problem, since, the value of is quite large, hence, the normal approximation to the binomial distribution can be used to calculate the required probabilities.

The binomial random variable, , will approximately follow the normal distribution with mean and variance .

Therefore, follows distribution.

Since, a discrete random variable is being approximated to a continuous random variable, hence, continuity correction needs to be used, that is:

The problem of interest is to calculate the probability that atleast 34 people out of those surveyed have blood group of O negative. In other words, , needs to be determined.

After applying the continuity correction, the problem becomes:

It is known that .

Now calculating the required probabilty using the normal distribution as shown below:

Let  . Therefore,

The value of the expression is calculated using the command "=NORMSDIST()" in MS-Excel. The screenshot is shown below:

This implies, .

Substituing the value of in the following equation:

Hence, it can be concluded that there is almost 14.06% chance that atleast 34 people out of the people investigated will possess O negative blood group.