in new york city the average response time for calls to the first depatrment is a bout 5 minutes with a standarrd deviation of 1.5 minuties suppose response follow a normal distribuation.
1-what is the probability that a response time exceeds 9 minutes/
2-compute the top 10 th percentile for response times ?
To answer these questions, we can use the properties of the normal distribution and the z-score formula. The z-score formula is given by:
z = (x - μ) / σ
Where:
- x is the observed value
- μ is the mean of the distribution
- σ is the standard deviation of the distribution
1. To find the probability that the response time exceeds 9 minutes, we need to calculate the area under the normal curve to the right of 9 minutes. In other words, we want to find P(X > 9), where X is the response time.
Step 1: Calculate the z-score for 9 minutes:
z = (9 - 5) / 1.5 = 2.67
Step 2: Find the probability corresponding to the z-score of 2.67 using a standard normal distribution table or a calculator with a normal distribution function. The probability will be the area to the right of the z-score.
Using a standard normal distribution table, the probability associated with a z-score of 2.67 is approximately 0.0038.
Therefore, the probability that a response time exceeds 9 minutes is approximately 0.0038 or 0.38%.
2. To compute the top 10th percentile for response times, we need to find the value such that only 10% of the distribution lies above it. In other words, we want to find the value, x, where P(X > x) = 0.10.
Step 1: Convert the percentile to a z-score by finding the z-score associated with 10% in the standard normal distribution table. The z-score will be negative since we are looking for a value to the right.
Using a standard normal distribution table, the z-score associated with 10% is approximately -1.28.
Step 2: Use the z-score formula to find the value, x:
-1.28 = (x - 5) / 1.5
Solving for x, we get:
x = -1.28 * 1.5 + 5 = 3.08
Therefore, the top 10th percentile for response times is approximately 3.08 minutes.