The average length of time, per week, that students at a certain university spend on homework is normally distributed with a mean of 24.3 hours and a standard deviation of 1.4 hours. Jane tells her parents she spends more time studying than 75% of the students on campus. How many hours per week must Jane spend on homework for this to be true?

Use normal distribution,

σ=1.4,
μ=24.3
Look up
z=Z-score for one-tail distribution for 0.75
Hours = μ+z*σ

Well, if Jane wants to spend more time studying than 75% of the students on campus, she better grab some extra coffee and put in some serious hours. According to the information given, we know that the average time is 24.3 hours and the standard deviation is 1.4 hours.

To find out how many hours Jane needs to study, we can use a z-table or a statistical calculator. Since we're dealing with a normal distribution, we need to find the z-score associated with the 75th percentile.

Luckily for us, the z-score corresponding to the 75th percentile is 0.674. Now we can use the formula z = (X - μ) / σ to solve for X, where X is the number of hours Jane needs to study.

So, plugging in the values, we have:
0.674 = (X - 24.3) / 1.4

Now we just need to solve for X:
0.674 * 1.4 = X - 24.3
0.9436 = X - 24.3
X = 0.9436 + 24.3
X ≈ 25.2436

So, Jane needs to spend approximately 25.2436 hours per week on homework to study more than 75% of the students on campus. Hope that helps! Now go grab some popcorn and cheer Jane on as she dives into her books.

To find out how many hours Jane must spend on homework per week, we need to find the z-score that corresponds to the 75th percentile of the normal distribution.

Step 1: Calculate the z-score using the formula:
z = (x - μ) / σ

Where:
x = the value we want to find (Jane's study time)
μ = mean of the distribution (24.3 hours)
σ = standard deviation of the distribution (1.4 hours)

Step 2: Look up the z-score on the z-table to find the corresponding percentile.

Step 3: Set up and solve the equation to find Jane's study time.

Let's do the calculations:

Step 1:
z = (x - μ) / σ
z = (x - 24.3) / 1.4

Step 2:
Looking up the z-score of 75th percentile on the z-table, we find that it is approximately 0.6745.

Step 3:
Setting up the equation:
0.6745 = (x - 24.3) / 1.4

Solving for x:
0.6745 * 1.4 = x - 24.3
0.9443 = x - 24.3
x = 24.3 + 0.9443
x = 25.2443

Therefore, Jane must spend approximately 25.2443 hours per week on homework to study more than 75% of the students on campus.

To find out how many hours per week Jane must spend on homework for this to be true, we need to find the value corresponding to the 75th percentile of the normal distribution of the average length of time spent on homework.

Here's how you can calculate it:

Step 1: Convert the given mean and standard deviation into z-scores.
The z-score formula is: z = (x - mean) / standard deviation

For Jane to spend more time studying than 75% of the students, we need to find the z-score associated with the 75th percentile. This z-score will give us the number of standard deviations above the mean Jane needs to be.

Step 2: Look up the z-score in the z-table.
The z-table provides the cumulative probability of a standard normal distribution up to a given z-score. By looking up the z-score corresponding to the 75th percentile, we can find the value that divides the lower 75% from the upper 25%.

Step 3: Convert the z-score back into the original units (hours).
Using the z-score value obtained in step 2, we can convert it back into the original units by multiplying it by the standard deviation and adding it to the mean.

Let's go through these steps to calculate the answer:

Step 1: Find the z-score corresponding to the 75th percentile.
To find the z-score, we use the equation:
z = (x - mean) / standard deviation

Substituting the given values:
75th percentile = 0.75
mean = 24.3 hours
standard deviation = 1.4 hours

z = (x - 24.3) / 1.4

Step 2: Look up the z-score in the z-table.
By consulting a standard normal distribution table or using a calculator with a normal distribution feature, we find that the z-score corresponding to the 75th percentile is approximately 0.674.

Step 3: Convert the z-score back into the original units.
Using the z-score value obtained in step 2, we can calculate Jane's required hours of study per week:

0.674 = (x - 24.3) / 1.4

Multiply both sides of the equation by 1.4 to isolate x:
0.674 * 1.4 = x - 24.3

0.9436 = x - 24.3

Add 24.3 to both sides of the equation to solve for x:
x = 24.3 + 0.9436

x ≈ 25.24

Therefore, Jane must spend approximately 25.24 hours per week on homework to spend more time studying than 75% of the students on campus.