Question

Use the given values of n and p to find the minimum usual value μ - 2σ and the maximum usual value μ + 2σ.

Answer 1

Answer:

$\mu -2\sigma = 915.428 - 2* 22.87=869.69$

$\mu +2\sigma = 915.428 + 2* 22.87=961.17$

And the best option would be:

4. Minimum: 869.69; maximum: 961.17

Step-by-step explanation:

We can assume that the variable of interst X is distributed with a binomial distribution and we can use the normal approximation.

For this case the mean would be given by:

$E(X) = np = 2136 *(\frac{3}{7})= 915.428$

And the standard deviation would be:

$\sigma = \sqrt{2136*(\frac{3}{7}) (1-\frac{3}{7})} =22.87$

And if we find the limits we got:

$\mu -2\sigma = 915.428 - 2* 22.87=869.69$

$\mu +2\sigma = 915.428 + 2* 22.87=961.17$

And the best option would be:

4. Minimum: 869.69; maximum: 961.17