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 = 31 minutes and standard deviation σ = 7 minutes.
(a) What is the probability that a first interview will last 40 minutes or longer? (Use 3 decimal places.)
(b) Two first interviews are usually scheduled per day. What is the probability that the average length of time for the two interviews will be 40 minutes or longer? (Use 3 decimal places.)
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.
If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events.
To answer these questions, we need to use the concept of the normal distribution and specifically the Z-score.
(a) In order to find the probability that a first interview will last 40 minutes or longer, we need to find the area under the normal curve from 40 minutes to positive infinity.
To do this, we first need to calculate the Z-score for 40 minutes. The Z-score measures how many standard deviations a given value is from the mean. It is calculated using the formula:
Z = (X - μ) / σ
Where:
X = 40 minutes (the value we are interested in)
μ = 31 minutes (mean)
σ = 7 minutes (standard deviation)
Plugging these values into the formula, we get:
Z = (40 - 31) / 7 = 1.286
Next, we need to find the probability associated with this Z-score. We can use a Z-table or a statistical calculator to find the corresponding probability. Looking up the Z-score of 1.286 in a Z-table, we find that the probability is approximately 0.900.
However, this probability is the cumulative probability up to 40 minutes. Since we are interested in the probability of 40 minutes or longer, we need to subtract this cumulative probability from 1. So:
Probability = 1 - 0.900 = 0.100
Therefore, the probability that a first interview will last 40 minutes or longer is 0.100 or 10% (rounded to 3 decimal places).
(b) Since we are dealing with the average length of time for two interviews, the Central Limit Theorem states that the distribution of the sample means will approach a normal distribution as the sample size increases, regardless of the shape of the population distribution.
For two first interviews, the mean of the average length is still 31 minutes. However, the standard deviation changes. According to the Central Limit Theorem, the standard deviation of the sample mean (also known as the standard error) can be calculated by dividing the population standard deviation by the square root of the sample size.
So, the standard deviation of the average length for two interviews would be:
σ_average = σ / sqrt(n)
Where:
σ = 7 minutes (standard deviation from the original problem)
n = 2 (number of interviews)
Plugging these values into the formula, we get:
σ_average = 7 / sqrt(2) ≈ 4.949
Now, we can calculate the Z-score for 40 minutes using the new standard deviation:
Z = (X - μ) / σ_average
Z = (40 - 31) / 4.949 ≈ 1.818
Using a Z-table or a statistical calculator, we find that the probability associated with a Z-score of 1.818 is approximately 0.964.
Therefore, the probability that the average length of time for the two interviews will be 40 minutes or longer is 0.964 or 96.4% (rounded to 3 decimal places).