Two times the least of three consecutive odd integers exceeds three times the greatest by 15

1 answer:

3 0

Three consecutive odd integers would be x, x+2, and x+4 assuming x is an odd integer. The smallest of these is x. Two times x is 2x. The greatest integer is x+4. Three times this is 3 (x+4), and if you distribute you get 3x+12. If 2x exceeds this by 15, you would make it 3x+12-15. If you add the like terms, 12+(-15) is -3. So, you have 2x=3x-3. Subtract 3x from both sides. 2x-3x is -1x, or -x. Now we have -x=-3. Divide by -, or -1 on both sides. Now we have x=3. You can substitute x for 3 for any of the consecutive odd integers to find their value.

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Step-by-step explanation:

I am going to use the normal approximation to the binomial to solve this question.

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:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

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:

$Z = \frac{X - \mu}{\sigma}$

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 $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .