. The average time a student spends studying just prior to a test is 36 minutes with a standard deviation of 7 minutes. This distribution approximates a normal curve. Find the percent of students who study just prior to the test for. a) more than 60 minutes b) between 55 and 59 minutes
To find the percent of students who study just prior to the test for more than 60 minutes, we need to calculate the z-score and then use the standard normal distribution table.
a) Let's calculate the z-score first:
z = (x - μ) / σ
where x is the value we want to find the percentage for, μ is the mean, and σ is the standard deviation.
Given:
x = 60 minutes
μ = 36 minutes
σ = 7 minutes
Calculating the z-score:
z = (60 - 36) / 7
z = 24 / 7
z ≈ 3.43
Using the standard normal distribution table or a calculator, we find that the area to the right of z = 3.43 is approximately 0.0003.
To convert this to a percentage, we multiply by 100:
Percentage = 0.0003 * 100 ≈ 0.03%
Therefore, the percent of students who study just prior to the test for more than 60 minutes is approximately 0.03%.
b) To find the percent of students who study between 55 and 59 minutes, we need to calculate the z-scores for both values and then find the difference in the cumulative areas.
Calculating the z-scores:
For 55 minutes:
z1 = (55 - 36) / 7
z1 ≈ 2.71
For 59 minutes:
z2 = (59 - 36) / 7
z2 ≈ 3.29
Using the standard normal distribution table or a calculator, we find the cumulative areas for z1 and z2:
Area for z1 ≈ 0.9961
Area for z2 ≈ 0.9993
To find the percentage between 55 and 59 minutes:
Percentage = (Area for z2) - (Area for z1)
Percentage ≈ 0.9993 - 0.9961
Percentage ≈ 0.0032
Therefore, the percent of students who study just prior to the test between 55 and 59 minutes is approximately 0.32%.
To find the percent of students who study just prior to the test for more than 60 minutes or between 55 and 59 minutes, we need to use the properties of the normal distribution.
a) To find the percent of students who study for more than 60 minutes, we can use the concept of the standard normal distribution. Let's first standardize the value of 60 minutes using the formula:
z = (x - μ) / σ
Where:
- x is the value we want to standardize (60 minutes in this case)
- μ is the mean of the distribution (36 minutes)
- σ is the standard deviation of the distribution (7 minutes)
Plugging in the values, we have:
z = (60 - 36) / 7
z = 24 / 7
z ≈ 3.43
Now, we need to find the area under the standard normal curve to the right of z = 3.43, which represents the percentage of students who study for more than 60 minutes. We can use a standard normal distribution table or a calculator to find this area. Using a standard normal distribution table, we find that the area to the right of z = 3.43 is approximately 0.0003.
Therefore, the percent of students who study just prior to the test for more than 60 minutes is approximately 0.03%, or 0.0003 as a decimal.
b) To find the percent of students who study between 55 and 59 minutes, we can standardize both values separately and then find the area between them. Let's first standardize the value of 55 minutes:
z1 = (55 - 36) / 7
z1 = 19 / 7
z1 ≈ 2.71
Next, let's standardize the value of 59 minutes:
z2 = (59 - 36) / 7
z2 = 23 / 7
z2 ≈ 3.29
Now, we need to find the area under the standard normal curve between z1 = 2.71 and z2 = 3.29. This represents the percentage of students who study between 55 and 59 minutes. Using a standard normal distribution table or a calculator, we find that the area between z1 = 2.71 and z2 = 3.29 is approximately 0.0012.
Therefore, the percent of students who study just prior to the test between 55 and 59 minutes is approximately 0.12%, or 0.0012 as a decimal.