Just before a referendum on a school budget, a local newspaper polls 354 voters to predict whether the budget will pass. Suppose the budget has the support of 55% of the voters. What is the probability that the newspaper's sample will lead it to predict defeat?

3.07% probability that the newspaper's sample will lead it to predict defeat

Step-by-step explanation:

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

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

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.

For a proportion p with n voters, we use

In this problem, we have that:

So

What is the probability that the newspaper's sample will lead it to predict defeat?

That is, a percentage of 50% or less, which is the pvalue of Z when X = 0.5. So

has a pvalue of 0.0307.

3.07% probability that the newspaper's sample will lead it to predict defeat

a negative exponent is translated as follows:

so, we have

1.3

step-by-step explanation:

3x2=6   9x+6

bring down the 6 on both sides

6+6=12

12= 9x

divide by 9

12/9 =

1.3 =x