7. Professors’ Salaries The average salary for a Queens

College full professor is $85,900. If the average salaries
are normally distributed with a standard deviation of
$11,000, find these probabilities. a. 0.3557 (TI: 0.3547)
a. The professor makes more than $90,000.
b. The professor makes more than $75,000.
Source: AAUP, Chronicle of Higher Education.

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a. Find the Z score

z score = (X value - the mean)/ Standard Deviation

(90,000 - 85,900)/11,000 = .372

Then find the probability of the Z score happening

P(Z > .372)

Note: Find the probability using a TI- 83/84 calculator.

Keys: 2nd, Vars, Normalcdf (Lower:.372 Upper: 10,000)

Final answer: .354

Well, let's crunch those numbers and have a little fun, shall we?

a. The professor makes more than $90,000:
To find this probability, we need to calculate the z-score first. The z-score can be calculated using the formula: z = (X - μ) / σ, where X is the value we're interested in, μ is the mean, and σ is the standard deviation.

Using the given values, we have:
z = (90,000 - 85,900) / 11,000
z = 0.3736

Now, we need to find the probability associated with a z-score of 0.3736. So, we consult our trusty normal distribution table and find that the probability is approximately 0.6480.

Therefore, the probability that a professor makes more than $90,000 is about 0.6480. Quite a boost for their pocketbooks!

b. The professor makes more than $75,000:
Similarly, let's calculate the z-score first:
z = (75,000 - 85,900) / 11,000
z = -0.9864

Now, we check the normal distribution table and find that the probability associated with a z-score of -0.9864 is approximately 0.1637.

Hence, the probability that a professor makes more than $75,000 is about 0.1637. They might not be rolling in dough, but they're doing alright!

I hope that answers your question with a dash of humor. If you have any more inquiries, I'm here to entertain!

To find the probabilities for these questions, we will use the normal distribution. The normal distribution is a bell-shaped curve that describes the distribution of values in a data set.

Step 1: Standardize the values
To use the normal distribution table or a calculator, we need to standardize the values by calculating the z-score.
The formula for calculating the z-score is:
z = (x - μ) / σ

Where:
x = the given value (salary)
μ = the mean (average salary)
σ = the standard deviation

For part a, the professor makes more than $90,000, so x = $90,000.
For part b, the professor makes more than $75,000, so x = $75,000.

Step 2: Find the z-scores
Using the formula above, we calculate the z-scores for each value.

For part a:
z = ($90,000 - $85,900) / $11,000

For part b:
z = ($75,000 - $85,900) / $11,000

Step 3: Look up the probabilities
Now, we can look up the probabilities corresponding to the calculated z-scores in the z-table or use a calculator.

For part a, we want to find the probability that the professor makes more than $90,000. Looking up the z-score from Step 2 in the z-table or using a calculator, we find that the corresponding probability is 0.3547 (TI: 0.3557).

For part b, we want to find the probability that the professor makes more than $75,000. Looking up the z-score from Step 2 in the z-table or using a calculator, we find the corresponding probability.

Note: The normal distribution table typically provides the probabilities for values less than the given z-score. However, because we want the probability that the professor makes MORE than the given values, we need to subtract the probability from 1.

To find the probability of making more than $75,000:
P(x > $75,000) = 1 - P(x < $75,000)

Step 4: Calculate the probabilities

For part b:
P(x > $75,000) = 1 - P(x < $75,000)

Substituting the calculated z-score into the formula, we find:
P(x > $75,000) = 1 - 0.9115 (TI: 1 - 0.9104) = 0.0885 (TI: 0.0896)

Therefore, the probabilities are:
a. The professor makes more than $90,000: 0.3557 (TI: 0.3547)
b. The professor makes more than $75,000: 0.0885 (TI: 0.0896)

Please note that the TI values may slightly differ due to rounding in the calculations.