The time required to finish a test in normally distributed with a mean of 80 minutes and a standard
deviation of 15 minutes. What is the probability that a student chosen at random will finish the test in more
than 110 minutes?
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.
To find the probability that a student chosen at random will finish the test in more than 110 minutes, we need to use the concept of the standard normal distribution.
1. Convert the given data to the standard normal distribution:
- Calculate the z-score using the formula: z = (x - μ) / σ, where x is the given value, μ is the mean, and σ is the standard deviation.
- Substitute the values: x = 110 minutes, μ = 80 minutes, and σ = 15 minutes.
- Calculate the z-score: z = (110 - 80) / 15 = 2.
2. Use the standard normal distribution table or a calculator to find the probability associated with the calculated z-score.
- The standard normal distribution table provides the probability of a value being less than a specific z-score.
- However, since we want to find the probability of the value being more than the given z-score, we need to subtract the probability obtained from 1 to get the desired probability.
If you have access to the standard normal distribution table:
- Look up the z-score we calculated (2) in the table.
- The table will provide the corresponding probability (P(Z < 2)).
- Subtract the obtained probability from 1 to find the probability of the value being more than 110 minutes.
If you have a calculator with built-in functionality for the standard normal distribution:
- Use the calculator's functionality to find the probability associated with the z-score 2.
- The calculator will provide the probability of the value being less than 110 minutes (P(Z < 2)).
- Subtract the obtained probability from 1 to find the probability of the value being more than 110 minutes.
By following these steps, you should be able to find the probability that a student chosen at random will finish the test in more than 110 minutes.