Question

question 9
the personnel office at a large electronics firm regularly schedules job interviews and maintains records of the interviews. from the
past records, they have found that the length of a first interview is normally distributed with mean μ = 35 minutes and standard
deviation σ = 7 minutes. enter your answer as a decimal rounded to 4 places.
a) what is the probability that a single interview will last 40 minutes or longer? (that is, ≥ 40 minutes)
enter answer
b) what is the probability that a sample of 9 interviews will have an average length of 40 minutes or longer?
enter answer

Explanation:

Step1: Calculate z - score for single interview

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\). Given \(\mu = 35\), \(\sigma=7\), and \(x = 40\).

$$z=\frac{40 - 35}{7}=\frac{5}{7}\approx0.7143$$

Step2: Find probability for single interview

Using the standard normal distribution table \(P(X\geq40)=1 - P(X < 40)\). From the standard normal table, \(P(Z<0.7143)\approx0.7625\). So \(P(X\geq40)=1 - 0.7625 = 0.2375\)

Step3: Calculate z - score for sample mean

The formula for the z - score of the sample mean \(\bar{x}\) is \(z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\). Given \(n = 9\), \(\mu = 35\), \(\sigma = 7\), and \(\bar{x}=40\)

$$z=\frac{40 - 35}{\frac{7}{\sqrt{9}}}=\frac{5}{\frac{7}{3}}=\frac{15}{7}\approx2.1429$$

Step4: Find probability for sample mean

Using the standard normal distribution table \(P(\bar{X}\geq40)=1 - P(\bar{X}<40)\). From the standard normal table, \(P(Z < 2.1429)\approx0.9839\). So \(P(\bar{X}\geq40)=1-0.9839 = 0.0161\)

Answer:

a) \(0.2375\)
b) \(0.0161\)