Question

Assume that X, the starting salary offer for sociology majors, is normally distributed with a mean of $43,256 and a standard deviation of $3,150. Use the following Distributions tool to help you answer the questions. (Note: To begin, click on the button in the lower left hand corner of the tool that displays the distribution and a single orange line.) The probability that a randomly selected sociology major received a starting salary offer less than $48,600 is . The probability that a randomly selected sociology major received a starting salary offer between $42,000 and $48,600 is . (Hint: The standard normal distribution is perfectly symmetrical about the mean, the area under the curve to the left (and right) of the mean is 0.5. Therefore, the area under the curve between the mean and a z-score is computed by subtracting the area to the left (or right) of the z-score from 0.5.) What percentage of sociology majors received a starting offer between $36,000 and $42,000? 66.61% 2.87% 97.13% 33.39%

Answer 1

Answer:

a) 95.54% or 0.9554

b) 61.08% or 0.6108

c) 33.39%

Step-by-step explanation:

It is given that the distribution of the salaries follow a normal distribution, so we will use the concept of z-scores to answer the given questions.

Mean salary = u = $ 43,256

Standard Deviation = $\sigma$ = $ ,3150

Part a) Probability that salary is less than $ 48,600

We have to find Probability (Salary < 48600). In symbolic form we can represent this by P(X < 48600).

We can solve this problem, by converting this value to corresponding z score, and using z table to find the said probability.

The formula of z scores is:

$z=\frac{x-u}{\sigma}$

Using the values, we get:

$z=\frac{48600-43256}{3150}=1.70$

So, P(X < 48600) is equivalent to P(z < 1.70). Using the z-table we can find the probability of z score being less than 1.70 i.e.

P(z < 1.70) = 0.9554

Since, P(X < 48600) = P(z < 1.70), we can write:

The probability that a randomly selected sociology major received a starting salary offer less than $48,600 is 0.9554 or 95.54%

Part b) Probability that the salary is between 42,000 and 48,600

We have to find P(42000 < X < 48600). We will use the same method as used in the previous part. First we have to convert the given values to z scores.

42000 converted to z score will be:

$z=\frac{42000-43256}{3150}=-0.40$

48600 converted to z score will be:

$z=\frac{48600-43256}{3150}=1.70$

So,

P(42000 < X < 48600) = P(-0.40 < z < 1.70)

According to the symmetry and properties of z distribution, we can write:

P(-0.40 < z < 1.70) = P(z < 1.70) - P(z < -0.40)

= 0.9554 - 0.3446

= 0.6108

Since, P(42000 < X < 48600) = P(-0.40 < z < 1.70), we can conclude:

The probability that a randomly selected sociology major received a starting salary offer between $42,000 and $48,600 is 0.6108 or 61.08%

Part c) Percentage of salaries between 36000 and 42000

We have to find P(36000 < X < 42000). This question is exactly similar to the previous one, just with the change of values. So again, first we find the z scores.

36000 converted to z score will be:

$z=\frac{36000-43256}{3150}=-2.30$

42000 converted to z score will be:

$z=\frac{42000-43256}{3150}=-0.40$

So,

P(36000 < X < 42000) = P(-2.30 < z < -0.40)

= P(z < -0.40) - P(z < -2.30)

= 0.3446 - 0.0107

= 0.3339

Since, P(36000 < X < 42000) = P(-2.30 < z < -0.40), we can conclude:

The percentage of sociology majors received a starting offer between $36,000 and $42,000 is 0.3339 (33.39%)