At a local restaurant, the amount of time that customers have to wait for their food is normally distributed with a mean of 24 minutes and a standard deviation of 3 minutes. What is the probability that a randomly selected customer will have to wait less than 25 minutes, to the nearest thousandth?
25 min is 1/3 s.d. above the mean ... z-score of .333...
use a z-score table to find the portion of the population
... below a 1/3 z-score
To find the probability that a randomly selected customer will have to wait less than 25 minutes, we need to calculate the area under the normal distribution curve to the left of 25 minutes.
Step 1: Standardize the value
To standardize 25 minutes, we need to calculate the z-score.
Z = (x - μ) / σ
Where:
x = 25 minutes (value we want to standardize)
μ = 24 minutes (mean)
σ = 3 minutes (standard deviation)
Z = (25 - 24) / 3
Z = 1 / 3
Step 2: Find the z-score in the standard normal distribution table
Using the standard normal distribution table (also known as the z-table), we can find the cumulative probability associated with the z-score. In this case, z = 1/3.
The table provides the cumulative probability to the left of the z-score. Based on the z-table, the cumulative probability associated with a z-score of 1/3 is approximately 0.630.
Step 3: Calculate the probability
The cumulative probability is the probability that a randomly selected customer will have to wait less than 25 minutes.
Probability = 0.630 (to the nearest thousandth)
Therefore, the probability that a randomly selected customer will have to wait less than 25 minutes is approximately 0.630.
To find the probability that a randomly selected customer will have to wait less than 25 minutes, we can use the standard normal distribution.
The first step is to convert the given values into z-scores. The formula for calculating the z-score is:
z = (x - μ) / σ
where:
z is the z-score,
x is the given value,
μ is the mean,
σ is the standard deviation.
In this case, the given value is 25 minutes, the mean is 24 minutes, and the standard deviation is 3 minutes.
z = (25 - 24) / 3
z = 1 / 3
Next, we need to find the cumulative probability associated with the z-score using a z-table or a statistical software. The cumulative probability represents the area under the normal curve up to the corresponding z-score.
Looking up the z-score of 1/3 in a standard normal distribution table, we find that the area to the left of this z-score is approximately 0.6293.
Therefore, the probability that a randomly selected customer will have to wait less than 25 minutes is approximately 0.6293, to the nearest thousandth.
Note: It is important to remember that z-scores are based on a standard normal distribution, which has a mean of 0 and a standard deviation of 1. To use the z-table, you may need to convert the given values to match this standard distribution.