According to a survey, 63% of the Scottish population has visited woodland in the previous year. Skeptical about this claim, you decide to take a simple random sample of 650 people in this population, and you ask them if they had visited woodland in the previous year. You find that 60% of the sample replied "yes" to your question.
Assuming that the original survey's 63% claim is correct, what is the approximate probability that less than 60% of the sample would report that they had visited woodland in the previous year?

5.71% probability that less than 60% of the sample would report that they had visited woodland in the previous year

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 sampling propotion p in a sample of size n, we have that

In this problem, we have that:

What is the approximate probability that less than 60% of the sample would report that they had visited woodland in the previous year?

This is the pvalue of Z when X = 0.6. So

has a pvalue of 0.0571

5.71% probability that less than 60% of the sample would report that they had visited woodland in the previous year

the sum of all the angles in a triangle are equal to 180

so you can add   the 3 given angles together to equal 180:

5y + 2y + 2y = 180

or since all three angles have the same letter, combine the 3 angles to equal 180:

9y = 180