A survey among freshman at a certain university revealed that the number of hours spent studying the week before final exams was normally distributed with mean 25 and standard deviation 15. A sample of 36 students was selected. What is the probability that the average time spent studying for the sample was between 26.2 and 30 hours studying?
Write only a number as you answer. Round to 4 decimal places.
Use same process, but calculate both Z scores first, then use the table.
To find the probability that the average time spent studying for the sample was between 26.2 and 30 hours studying, we need to calculate the z-scores and then use the z-table.
First, find the z-score for 26.2 hours:
z1 = (26.2 - 25) / (15 / √36)
z1 = (26.2 - 25) / (15 / 6)
z1 = (26.2 - 25) / 2.5
z1 = 1.2 / 2.5
z1 = 0.48
Next, find the z-score for 30 hours:
z2 = (30 - 25) / (15 / √36)
z2 = (30 - 25) / (15 / 6)
z2 = (30 - 25) / 2.5
z2 = 5 / 2.5
z2 = 2
Now, look up these z-scores in the z-table. The probability between these two z-scores represents the probability that the average time spent studying for the sample was between 26.2 and 30 hours.
Using the z-table or a calculator, we find that the area to the left of z = 0.48 is 0.6844, and the area to the left of z = 2 is 0.9772.
Therefore, the probability that the average time spent studying for the sample was between 26.2 and 30 hours studying is:
0.9772 - 0.6844 = 0.2928
Rounded to 4 decimal places, the probability is approximately 0.2928.
To find the probability that the average time spent studying for the sample was between 26.2 and 30 hours, we need to standardize the values using the Central Limit Theorem and then use the Z-table.
First, we need to calculate the standard deviation of the sample mean. The standard deviation of the sample mean (also known as the standard error) is equal to the population standard deviation divided by the square root of the sample size. In this case, the population standard deviation is 15, and the sample size is 36.
Standard error (SE) = 15 / √36
SE = 15 / 6
SE = 2.5
Next, we need to standardize the values using the formula for Z-score:
Z = (X - μ) / SE
Where:
X = The value we are standardizing (in this case, 26.2 and 30 hours)
μ = The population mean (given as 25)
SE = The standard error we calculated previously (2.5)
For 26.2 hours:
Z1 = (26.2 - 25) / 2.5
Z1 = 1.2 / 2.5
Z1 = 0.48
For 30 hours:
Z2 = (30 - 25) / 2.5
Z2 = 5 / 2.5
Z2 = 2
Now, we can use the Z-table to find the probabilities associated with these Z-scores.
The Z-table gives us the area under the standard normal distribution curve. We want to find the probability between these two Z-scores, which represents the area between these values.
Using the Z-table, we look up the probabilities corresponding to Z1 = 0.48 and Z2 = 2.
The probability associated with Z1 = 0.48 is approximately 0.6844.
The probability associated with Z2 = 2 is approximately 0.9772.
To find the probability between these two Z-scores, we subtract the smaller probability from the larger probability.
Probability = 0.9772 - 0.6844
Probability = 0.2928
Therefore, the probability that the average time spent studying for the sample was between 26.2 and 30 hours is approximately 0.2928.