Let X denote the time in hours needed to locate and correct a problem in the software that governs the timing of traffic lights in the downtown area of a large city. Assume that X is normally distributed with mean 10 hours and variance 9.
(a) Find the probability that the next problem will require at most 15 hours to find and correct.
(b) The fastest 5% of repairs take at most how many hours to complete?
How do we do part b only using standard normal table?
standard deviation is the √ of the variance
the fastest 5% are more than 2 s.d. below the mean
How do we get that : 2 s.d below mean?
To answer part (b) using a standard normal table, we need to convert the given distribution to a standard normal distribution. This process involves finding the z-score corresponding to the desired probability.
The z-score is calculated using the formula:
z = (X - μ) / σ
Where:
- X is the value of interest (in this case, the number of hours)
- μ is the mean of the distribution
- σ is the standard deviation of the distribution
Since we are given the mean (μ = 10) and variance (σ^2 = 9), we can calculate the standard deviation (σ) by taking the square root of the variance: σ = √9 = 3.
Next, we need to find the z-score corresponding to the fastest 5% of repairs. This means finding the z-score that represents the value below which 5% of the data falls. In other words, we want to find the z-score such that the probability of being below that value is 0.05.
To determine the z-score from a standard normal table, we look for the closest probability value to 0.05 in the body of the table. In this case, the closest value is 0.0505.
Looking up this probability value in the table, we find that the corresponding z-score is approximately -1.645.
Finally, we can calculate the value of X (the number of hours) by rearranging the z-score formula:
X = μ + z * σ
Substituting the values, we get:
X = 10 + (-1.645) * 3
X = 10 - 4.935
X ≈ 5.065
Therefore, the fastest 5% of repairs take at most approximately 5.065 hours to complete.
sorry ... my mistake
... approx. 95% of the population lies within 2 s.d. of the mean
... but we want the bottom 5%
s.d. = √(variance) = √9 = 3
looking at a z-score table
... the bottom 5% is approx. 1.64 s.d. below the mean
1.64 * 3 hr =?