Question

Salaries for senior Civil engineers per year have known mean of 78 K dollars and standard deviation 5 K dollars. Provided that the salaries per year for senior Civil engineers are normally distributed, what's the probability of finding a Civil engineer whose salary is between 83 K and 93 K dollars? Find the answer without using the LSND program. (Write the answer in decimals)

Answer 1

Solution

mean = 78

standard deviation = 5

for x= 83

z = (83-78)/5 = 1.00

for x = 93

z = (93-78)/5 = 3.00

p(83<x<93) = p(1.00<z<3.00) = 0.1573

answer: 0.1573

Answer 2

To find the probability of a senior civil engineer's salary being between $83,000 and $93,000, we can use the properties of the normal distribution. Specifically, we need to calculate the area under the normal curve between these two salary values.

Step 1: Standardize the values

We'll first standardize the two salary values using the z-score formula:

z = (X - μ) / σ

Where: X = Salary value μ = Mean of the distribution (78,000 dollars) σ = Standard deviation of the distribution (5,000 dollars)

For the salary $83,000: z1 = (83,000 - 78,000) / 5,000 z1 = 1

For the salary $93,000: z2 = (93,000 - 78,000) / 5,000 z2 = 3

Step 2: Look up the probabilities

Next, we need to find the probabilities associated with these z-scores. We can use a standard normal distribution table or a calculator to find these values.

The probability of finding a civil engineer with a salary less than $83,000 is the same as finding the area to the left of the z-score of 1.

P(X < 83,000) = P(Z < 1) ≈ 0.8413 (from the standard normal distribution table)

The probability of finding a civil engineer with a salary less than $93,000 is the same as finding the area to the left of the z-score of 3.

P(X < 93,000) = P(Z < 3) ≈ 0.9987 (from the standard normal distribution table)

Step 3: Calculate the probability between the two salaries

To find the probability of a salary being between $83,000 and $93,000, we subtract the probability at z1 from the probability at z2:

P(83,000 < X < 93,000) = P(Z < 3) - P(Z < 1) ≈ 0.9987 - 0.8413 ≈ 0.1574

So, the probability of finding a civil engineer whose salary is between $83,000 and $93,000 is approximately 0.1574, or 15.74% (in decimals).

Answer 3

To find the probability of finding a senior Civil engineer whose salary is between 83 K and 93 K dollars, we can use the properties of a normal distribution and standardize the values using z-scores. The formula for calculating the z-score is:

z = (X - μ) / σ

where: X = the value we want to find the probability for (in this case, 83 K and 93 K dollars) μ = the mean of the distribution (78 K dollars) σ = the standard deviation of the distribution (5 K dollars)

Let's calculate the z-scores for the given values:

For X = 83 K dollars: z1 = (83 - 78) / 5 z1 = 1

For X = 93 K dollars: z2 = (93 - 78) / 5 z2 = 3

Now, we need to find the probability of the z-scores falling between 1 and 3. For a standard normal distribution (mean = 0, standard deviation = 1), we can use a z-table to find the probabilities.

Using the z-table, the probability of finding a z-score between 1 and 3 is approximately 0.1574.

Therefore, the probability of finding a senior Civil engineer whose salary is between 83 K and 93 K dollars is approximately 0.1574 (rounded to four decimal places).