Local food distributor claims that 18% of all food sold are vegetables. If 112 food items are sold are randomly selected, what is the probability that the proportion of vegetables sold differs from the population proportion by more than 2%?
I am trying to plug in the formula for standard score for a sample proportion, but I'm not getting anything that makes sense. I tried by hand and with my ti-84 but I'm not sure how to put it in the calf correctly. I like to use the calc to check my work even when I do it by hand.
To solve this problem, we need to calculate the probability that the proportion of vegetables sold differs from the population proportion by more than 2%.
First, we'll calculate the standard deviation of the sampling distribution of proportions. The formula for the standard deviation of a sample proportion is:
σ = sqrt((p * (1 - p)) / n)
Where:
σ = standard deviation of sample proportion
p = population proportion
n = sample size
In this case, the population proportion of vegetables sold is given as 18%, which can be expressed as 0.18. The sample size is 112. Plugging these values into the formula, we get:
σ = sqrt((0.18 * (1 - 0.18)) / 112)
Next, we calculate the margin of error, which is the maximum difference allowed between the sample proportion and the population proportion. In this case, we want to find the probability of the proportion differing by more than 2%, so the margin of error is 0.02.
Now, we can calculate the z-score using the formula:
z = (p ̂ - p) / σ
Where:
z = z-score
p ̂ = sample proportion
p = population proportion
σ = standard deviation of sample proportion
To find the probability that the proportion differs by more than 2%, we need to calculate both the probability on the left tail and the right tail. Since the standard normal distribution is symmetrical, we can calculate the probability on only one tail and then multiply it by 2 at the end.
Using the z-score, we can look up the probability in the standard normal distribution table or use a calculator to find the cumulative probability. If you are using a TI-84 calculator, follow these steps:
1. Press the "2nd" button, then press "VARS" to enter the "DISTR" menu.
2. Scroll down to find "normalcdf(", then press "Enter".
3. Enter the lower limit as -9999 (negative infinity), and the upper limit as the z-score obtained.
4. Press "Enter" to get the cumulative probability on the left tail.
5. Subtract this value from 1 to get the cumulative probability on the right tail.
6. Multiply the right tail probability by 2 to get the probability of the proportion differing by more than 2%.
Remember to use a negative sign in the z-score calculation if the sample proportion is less than the population proportion.
By following these steps, you should be able to use the TI-84 calculator correctly to solve this problem.