the average starting salary of this yaer's graduates of a large university (LU)is $25000 with a standard deviation of $5000.futhermore,it is known that the starting salaries are normally distributed.

a)what is the probability that a randomly selected LU graduate will have a starting of at least $31000?
b)individuals with starting salaries of less than $12200 receive a low income tax break.what percentage of the graduates will receive the tax break?
C)what are the minimum and the maximum starting salaries of the middle 95% of the LU graduates?
d)if 68 of the recent gradutes have salaries of at least $35600 , how many students graduated this year from this university>

a, b) Z = (score-mean)/SD

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions related to the Z scores.

c) Use same table to find percent (47.5%) between Z score and mean. Insert Z score in above equation and solve.

d) Find Z score, use table to get %. That % of total = 68.

a) To find the probability that a randomly selected LU graduate will have a starting salary of at least $31000, we can use the Z-score formula.

First, let's calculate the Z-score:
Z = (X - μ) / σ

Where:
X = Value we are interested in (starting salary of at least $31000)
μ = Mean (average starting salary of $25000)
σ = Standard deviation ($5000)

Z = (31000 - 25000) / 5000
Z = 6000 / 5000
Z = 1.2

Next, we consult the Z-table to find the probability corresponding to a Z-score of 1.2. The Z-table provides the cumulative probability up to a given Z-score. However, we are interested in the probability from the given Z-score (1.2) to the right (since we want a starting salary of at least $31000).

From the Z-table, we find that the cumulative probability (from the left) for a Z-score of 1.2 is approximately 0.8849. Therefore, the probability from 1.2 to the right is:

Probability = 1 - 0.8849
Probability = 0.1151 or 11.51%

So, the probability that a randomly selected LU graduate will have a starting salary of at least $31000 is 11.51%.

b) To find the percentage of graduates who will receive the low-income tax break (starting salaries less than $12200), we need to calculate the Z-score for $12200 and then determine the cumulative probability from the left in the Z-table.

Z = (X - μ) / σ
Z = (12200 - 25000) / 5000
Z = -12800 / 5000
Z = -2.56

From the Z-table, we find that the cumulative probability (from the left) for a Z-score of -2.56 is approximately 0.0054. Therefore, the percentage of graduates who will receive the tax break is:

Percentage = 0.0054 * 100
Percentage ≈ 0.54%

So, approximately 0.54% of the graduates will receive the tax break.

c) The middle 95% of the LU graduates corresponds to the range covered by two standard deviations on each side of the mean. To find the minimum and maximum starting salaries for this range, we need to calculate the Z-scores.

For the lower boundary:
Z = -1.96 (Z-score for the left 2.5% of the distribution, since we're looking for the middle 95%)
Minimum salary = μ + (Z * σ)
Minimum salary = 25000 + (-1.96 * 5000)
Minimum salary ≈ 15020

For the upper boundary:
Z = 1.96 (Z-score for the right 2.5% of the distribution, since we're looking for the middle 95%)
Maximum salary = μ + (Z * σ)
Maximum salary = 25000 + (1.96 * 5000)
Maximum salary ≈ 34980

Therefore, the minimum starting salary for the middle 95% of the LU graduates is approximately $15020, and the maximum starting salary is approximately $34980.

d) To estimate the number of students who graduated this year from the university, we'll assume that the number of graduates follows the same distribution as the starting salaries, which is normally distributed.

Given that 68 graduates have salaries of at least $35600, we can use the Z-score formula to find the Z-score corresponding to this value.

Z = (X - μ) / σ
Z = (35600 - 25000) / 5000
Z = 10600 / 5000
Z = 2.12

From the Z-table, we find that the cumulative probability (from the left) for a Z-score of 2.12 is approximately 0.9821, or 98.21%.

The 68 graduates represent approximately 98.21% of the total number of graduates.

Let's denote the total number of graduates as N. We can set up a proportion:

68 / N = 0.9821

Cross-multiplying, we have:

68 ≈ 0.9821 * N
N ≈ 68 / 0.9821
N ≈ 69.27

Therefore, we estimate that approximately 69 students graduated this year from the university.

a) To find the probability that a randomly selected LU graduate will have a starting salary of at least $31000, we need to calculate the z-score for this value and then find the area under the normal distribution curve to the right of that z-score.

First, we calculate the z-score using the formula:
z = (x - μ) / σ
where x is the desired starting salary, μ is the mean starting salary, and σ is the standard deviation.

For this problem:
x = $31000
μ = $25000
σ = $5000

z = ($31000 - $25000) / $5000
z = 6000 / 5000
z = 1.2

Next, we use a z-score table or a calculator to find the probability associated with this z-score. In this case, we want to find the area to the right of 1.2. The probability can be found as 1 - probability to the left of 1.2.

Using a z-score table, we can find that the probability to the left of 1.2 is approximately 0.8849. Therefore, the probability to the right of 1.2 is 1 - 0.8849 = 0.1151, or 11.51%.

So, the probability that a randomly selected LU graduate will have a starting salary of at least $31000 is approximately 11.51%.

b) To find the percentage of graduates who will receive the low income tax break (starting salary less than $12200), we need to calculate the z-score for this value and then find the area under the normal distribution curve to the left of that z-score.

Using the same formula as in (a):
z = ($12200 - $25000) / $5000
z = -12800 / $5000
z = -2.56

Looking up this z-score in a z-score table, we can find that the probability to the left of -2.56 is approximately 0.0051. Therefore, the percentage of graduates who will receive the tax break is 0.0051 * 100% = 0.51%.

So, approximately 0.51% of the graduates will receive the low income tax break.

c) To find the minimum and maximum starting salaries of the middle 95% of the LU graduates, we need to find the z-scores that correspond to the lower and upper percentiles of the distribution.

The middle 95% represents 100% - 95% = 5% on each tail of the normal distribution. We need to find the z-scores that correspond to 2.5% on each tail.

Using a z-score table, we can find that the z-score that corresponds to 2.5% is approximately -1.96.

With this z-score, we can calculate the values of the minimum and maximum starting salaries:
Minimum = μ + z * σ
= $25000 + (-1.96) * $5000
= $16120

Maximum = μ + z * σ
= $25000 + (1.96) * $5000
= $33880

So, the minimum starting salary of the middle 95% of the graduates is approximately $16120, and the maximum starting salary is approximately $33880.

d) If 68 of the recent graduates have salaries of at least $35600, we can use this information to estimate the total number of graduates for this year from the university.

First, we calculate the z-score for this value:
z = ($35600 - $25000) / $5000
z = $10600 / $5000
z = 2.12

Given that the 68 graduates represent 68%, we can use this information to find the total number of graduates:
Total graduates = (68 / 0.68) * 100%
= 100

Therefore, the estimated number of students who graduated this year from the university is 100.