Let Xb be the number of reservations that are accommodated. Xb has the binomial distribution with n trials and success probability p = 0.94
In general, if X has the binomial distribution with n trials and a success probability of p then P[Xb = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x) for values of x = 0, 1, 2, ..., n P[Xb = x] = 0 for any other value of x.
To use the normal approximation to the binomial you must first validate that you have more than 10 expected successes and 10 expected failures. In other words, you need to have n * p > 10 and n * (1-p) > 10.
Some authors will say you only need 5 expected successes and 5 expected failures to use this approximation. If you are working towards the center of the distribution then this condition should be sufficient. However, the approximations in the tails of the distribution will be weaker espeically if the success probability is low or high. Using 10 expected successes and 10 expected failures is a more conservative approach but will allow for better approximations especially when p is small or p is large.
If Xb ~ Binomial(n, p) then we can approximate probabilities using the normal distribution where Xn is normal with mean μ = n * p, variance σ² = n * p * (1-p), and standard deviation σ
I have noted two different notations for the Normal distribution, one using the variance and one using the standard deviation. In most textbooks and in most of the literature, the parameters used to denote the Normal distribution are the mean and the variance. In most software programs, the standard notation is to use the mean and the standard deviation.
The probabilities are approximated using a continuity correction. We need to use a continuity correction because we are estimating discrete probabilities with a continuous distribution. The best way to make sure you use the correct continuity correction is to draw out a small histogram of the binomial distribution and shade in the values you need. The continuity correction accounts for the area of the boxes that would be missing or would be extra under the normal curve.
Try this option: 1. Odd numbers, which are multiple of 15 and less then 100, are: 15, 45 and 75. 2. the sums of digits are: for 15 is 6, for 45 is 9, for 75 is 12. 3. this number is 75.
The number is 1. 1 multiplied by 100 gives 100 as answer. Lisa claims that <span>when multiplying any number between 0 and 10 by 100, the product is greater than 100. In this case the answer is equal to 100 not greater than 100.
So this proof shows that Lisa's claim is not correct. </span>
