153k views
2 votes
An opinion poll asks an SRS of 501 teenagers, "Generally speaking, do you approve or disapprove of legal gambling or betting? Suppose that, in fact, exactly 50% of all teenagers would say Approve if asked. (This is close to what polls show to be true.) The poll's statisticians tell us that the sample proportion who say Approve will vary in repeated samples according to a Normal distribution with mean 0.5 and standard deviation about 0.022. Using the 68-95-99.7 rule, what is the probability that fewer than 54.4% say Yes? (Use decimal notation. Give your answer as an exact proportion.) 97.5 probability: Inc

User JordanBean
by
8.4k points

2 Answers

4 votes

Final answer:

To find the probability that fewer than 54.4% say Approve in the survey, we can calculate the z-score and use the standard normal distribution table. The probability is practically 1, or 100%.

Step-by-step explanation:

To find the probability that fewer than 54.4% say Approve, we need to calculate the z-score for the given percentage and then use the z-score to find the probability from the standard normal distribution table.

The z-score is calculated as (x - mean) / standard deviation. In this case, x = 54.4%, mean = 50%, and standard deviation = 0.022. Plugging these values in, we get z = (54.4% - 50%) / 0.022 = 0.4 / 0.022 = 18.18.

Using the standard normal distribution table, we can find that the probability of getting a z-score less than 18.18 is practically 1. So the probability that fewer than 54.4% say Yes is practically 1, or 100%.

User Ajith Antony
by
8.0k points
2 votes

Final answer:

The question tells us that the sample proportion of teenagers who approve of legal gambling or betting follows a normal distribution with a mean of 0.5 and a standard deviation of 0.022. The probability that fewer than 54.4% of teenagers say yes is low.

Step-by-step explanation:

To solve this problem, we can use the concept of the standard normal distribution. The question tells us that the sample proportion of teenagers who approve of legal gambling or betting follows a normal distribution with a mean of 0.5 and a standard deviation of 0.022.

We want to find the probability that fewer than 54.4% of teenagers say yes.

We can convert this into a standard normal distribution by subtracting the mean and dividing by the standard deviation. The standardized value for 54.4% is (54.4% - 50%)/0.022 = 4.4/0.022 = 200.

Using the 68-95-99.7 rule, we know that 95% of the values fall within 2 standard deviations of the mean.

Since we are looking for values that are less than 54.4%, which is more than 2 standard deviations above the mean, the probability would be close to 0. Therefore, the probability that fewer than 54.4% say yes is very low.

User Beamer
by
8.6k points

No related questions found