suppose that the times taken to complete an online test are normally distributed with a mean of 45 minutes and a standard deviation of 12 minutes. find the probability
a. that for a randomly selected test, the time taken was more than 1 hour.
b. that the mean number of minutes taken by a random sample of 24 tests is less than 50 minutes
In order to solve these probability questions, we need to use the concept of the standard normal distribution. The standard normal distribution is a normal distribution with a mean of 0 and a standard deviation of 1.
To find the probabilities for the given questions, we need to standardize the values using z-scores. The formula for calculating the z-score is:
z = (x - μ) / σ
Where:
- z is the z-score
- x is the value we want to standardize
- μ is the mean of the distribution
- σ is the standard deviation of the distribution
Now, let's solve each question separately:
a. To find the probability that the time taken for a randomly selected test is more than 1 hour (which is 60 minutes), we need to find the z-score for 60 minutes and calculate the area under the standard normal distribution curve to the right of that z-score.
z = (60 - 45) / 12 = 15 / 12 = 1.25
To find the probability, we can use a standard normal distribution table or a calculator. Looking up the z-score of 1.25 in the table, we find that the area to the right of 1.25 is approximately 0.1056.
Therefore, the probability that the time taken for a randomly selected test is more than 1 hour is approximately 0.1056, or 10.56%.
b. To find the probability that the mean number of minutes taken by a random sample of 24 tests is less than 50 minutes, we need to find the z-score for 50 minutes and calculate the area under the standard normal distribution curve to the left of that z-score.
z = (50 - 45) / (12 / sqrt(24)) = 5 / (12 / sqrt(24)) = 1.29
Again, using a standard normal distribution table or a calculator, we find that the area to the left of 1.29 is approximately 0.9015.
Therefore, the probability that the mean number of minutes taken by a random sample of 24 tests is less than 50 minutes is approximately 0.9015, or 90.15%.
a. Z = (score-mean)/SD
b. Z = (score-mean)/SEm
SEm = SD/√n
For each, 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.