The mean starting salary for graduates in the spring of 2004 was $36,280. Assume that the distribution of starting salaries follows the normal distribution with a standard deviation of $3,300. What percent of the graduates have starting salaries between $35,000 and $40,000?

a. 5225
b. .0041
c. .1251
d. None of the above

Copy and paste the question into Google and the answer is there. Looks like it will be d, but check it out yourself

Here is the math

(35,000<X<40,000)=P((35,000-36,280)/3,300<(X-36,280)/3,300<(40,000-36,280)/3,300)=P(-0.3879<Z<1.1273)
=P(Z<1.1273)-P(Z<-0.3879)=0.8702-0.3491= 0.5211

To find the percentage of graduates with starting salaries between $35,000 and $40,000, we need to calculate the z-scores for each salary and then use the standard normal distribution table.

We can calculate the z-scores using the formula:

z = (x - μ) / σ

Where:
x = salary
μ = mean starting salary = $36,280
σ = standard deviation = $3,300

For $35,000:
z1 = (35,000 - 36,280) / 3,300

For $40,000:
z2 = (40,000 - 36,280) / 3,300

Calculating these values:
z1 = -0.3848
z2 = 1.1212

Now, we need to look up the respective z-scores in the standard normal distribution table.

Looking up the z1 value of -0.3848 in the table, we find the corresponding area under the curve is 0.3520.

Looking up the z2 value of 1.1212 in the table, we find the corresponding area under the curve is 0.8686.

To find the percentage between these two z-scores, we subtract the smaller area from the larger area and multiply by 100:

Percentage = (0.8686 - 0.3520) * 100 = 51.66%

Therefore, the correct answer is not given in the options (d. None of the above) as the percentage is not one of the provided choices. The percentage of graduates with starting salaries between $35,000 and $40,000 is approximately 51.66%.

To find the percent of graduates with starting salaries between $35,000 and $40,000, we need to calculate the z-scores for those values and use the z-table to find the corresponding percentiles.

First, let's calculate the z-scores:
Z1 = (X1 - mean) / standard deviation
Z2 = (X2 - mean) / standard deviation
where X1 is $35,000, X2 is $40,000, mean is $36,280, and standard deviation is $3,300.

Z1 = (35,000 - 36,280) / 3,300
Z2 = (40,000 - 36,280) / 3,300

Next, we need to use the z-table to find the percentiles corresponding to these z-scores.

Looking up the z-scores in the z-table, we find:
Z1 = -0.3909
Z2 = 1.1255

The z-table gives the area under the normal distribution curve to the left of a given z-score. Since we are interested in the area between Z1 and Z2, we calculate the area to the left of Z2 and subtract the area to the left of Z1.

Area between Z1 and Z2 = Area to the left of Z2 - Area to the left of Z1

From the z-table, we find the following values:
Area to the left of Z1 = 0.6480
Area to the left of Z2 = 0.8708

Area between Z1 and Z2 = 0.8708 - 0.6480 = 0.2228

Finally, we convert the decimal to a percentage:
0.2228 * 100 = 22.28%

So, the percent of graduates with starting salaries between $35,000 and $40,000 is approximately 22.28%.

None of the answer choices provided (a, b, c) matches the calculated value. Therefore, the correct answer is d) None of the above.