Question

Agnes Hammer is a senior majoring in management science. She has been interviewing with several companies for a job when she graduates, and she is curious about what starting salary offers she might receive. There are 140 seniors in the graduating class for her major, and more than half have received job offers. She asked 12 of her classmates at random what their annual starting salary offers were, and she received the following responses: $28,500 $35,500 $32,600 $36,000 $34,000 $25,700 $27,500 $29,000 $24,600 $31,500 $34,500 $26,800 Assume that starting salaries are normally distributed. Compute the mean and standard deviation for these data and determine the probability that Agnes will receive a salary offer of less than $27,000.

Answer 1

Answer:

Mean = 30516.67

Standard deviation, s = 3996.55

P(x < 27000) = 0.0011518

Step-by-step explanation:

Given the data:

28500 35500 32600 36000 34000 25700 27500 29000 24600 31500 34500 26800

Mean, xbar = Σx / n = 366200 /12 = 30516.67

Standard deviation, s = [√Σ(x - xbar) / n-1]

Using calculator, s = 3996.55

The ZSCORE = (x - mean) / s/√n

Zscore = (27000 - 30516.67) / (3996.55/√12)

Zscore = - 3516.67 / 1153.7046

Zscore = - 3.048

P(x < 27000) = P(Z < - 3.049) = 0.0011518