1. The time it takes a student to cycle to school is normally distributed with a mean of 15 minutes and a variance of 2. The student has to be at school at 8.00a.m. At what time should the student leave her house so that she will be late only 4% of the time?
2. In a study for Health Statistics, it was found that the mean height of female 20-25 years old was 64.1 inches. If height is normally distributed with a standard deviation of 2.8 inches.
a)Determine the height required to be in the top 10% of all 20-25 years old females.
b)A “one-size-fits all” robe is being designed that should fit 90% of 20-25 years old females; what heights constitute the middle 90% of all 20-25 years old females?
3. The compressive strength of cement is assumed to be normally distributed with a mean of 8000 kilograms per sq cm and a standard deviation of 200 kilograms per sq. cm.
a)Find the probability that the strength is less than 8200 kg per sq. cm
b)Find the probability that the strength is between 7500 and 7700 kg per sq. cm
c)Determine the value for which the probability that the strength of cement is below this value is 90%.
1. Z = (score-mean)/SD
Variance = SD^2
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the Z score related to that proportion. Remember that the Z score will be negative. Plug in the values in the above equation and solve for the score.
2 & 3. Follow a similar processes for these values.
1. To solve this problem, we need to find the time at which the student should leave her house so that she will be late only 4% of the time.
Given that the time it takes to cycle to school follows a normal distribution with a mean of 15 minutes and a variance of 2, we can standardize this distribution using the z-score formula:
z = (x - μ) / σ
Where:
z is the z-score
x is the value we want to find (departure time)
μ is the mean (15 minutes)
σ is the standard deviation (√2)
Using a z-table or a calculator with a cumulative probability function, we can find the z-value corresponding to the 4th percentile (since the student wants to be late only 4% of the time) which is -1.75.
Now, we can rearrange the z-score formula to solve for x:
-1.75 = (x - 15) / √2
Solving for x:
x - 15 = -1.75 * √2
x = 15 - (1.75 * √2)
This gives us the departure time the student should leave her house to be late only 4% of the time.
2. a) To find the height required to be in the top 10% of all 20-25 years old females, we need to determine the z-score corresponding to the 90th percentile.
Using the z-score formula:
z = (x - μ) / σ
Given that the mean height is 64.1 inches and the standard deviation is 2.8 inches, we can solve for x:
z = (x - 64.1) / 2.8
To find the z-score for the 90th percentile, we can use a z-table or a calculator with the inverse cumulative probability function. The z-score corresponding to the 90th percentile is approximately 1.28.
Now, we can solve for x:
1.28 = (x - 64.1) / 2.8
Solving for x:
x - 64.1 = 1.28 * 2.8
x = 64.1 + (1.28 * 2.8)
This gives us the height required to be in the top 10% of all 20-25 years old females.
b) To determine the heights that constitute the middle 90% of all 20-25 years old females (thus fitting the one-size-fits-all robe), we need to find the z-scores for the 5th and 95th percentiles.
Using the z-score formula:
z = (x - μ) / σ
First, find the z-score corresponding to the 5th percentile:
z1 = (x1 - 64.1) / 2.8
Then, find the z-score corresponding to the 95th percentile:
z2 = (x2 - 64.1) / 2.8
Using a z-table or a calculator with the inverse cumulative probability function, we can find that the z-scores for the 5th and 95th percentiles are approximately -1.65 and 1.65, respectively.
Now, we can solve for x1 and x2:
-1.65 = (x1 - 64.1) / 2.8
x1 = (-1.65 * 2.8) + 64.1
1.65 = (x2 - 64.1) / 2.8
x2 = (1.65 * 2.8) + 64.1
These heights (x1 and x2) represent the range of heights that constitute the middle 90% of all 20-25 years old females.
3. a) To find the probability that the strength of cement is less than 8200 kg per sq. cm, we need to calculate the z-score and use a z-table or a calculator with a cumulative probability function.
Given that the mean compressive strength is 8000 kg per sq. cm and the standard deviation is 200 kg per sq. cm, we can solve for the z-score:
z = (x - μ) / σ
z = (8200 - 8000) / 200
Now, using a z-table or a calculator with a cumulative probability function, we can find the probability corresponding to the z-score.
b) To find the probability that the strength is between 7500 and 7700 kg per sq. cm, we can follow a similar process as in part (a).
Calculate the z-scores for 7500 and 7700 using the formula:
z = (x - μ) / σ
Then, find the cumulative probabilities for each z-score and subtract the smaller probability from the larger probability to get the probability of the strength being between the two values.
c) To determine the value for which the probability that the strength of cement is below this value is 90%, we need to find the z-score corresponding to the 90th percentile.
Using the z-score formula:
z = (x - μ) / σ
Solve for x by rearranging the formula:
x = (z * σ) + μ
Calculate the z-score for the 90th percentile and then substitute it back into the formula to find the corresponding value of x.