Math Problem Statement
solve all parts completely
Solution
We are given that the time customers spend waiting at the cashier follows a normal distribution with a mean of 15 minutes and a standard deviation of 2 minutes. We are tasked with finding the probabilities for three different scenarios.
Part (i): Probability that the waiting time is more than 20 minutes
To calculate this, we use the z-score formula:
Where:
- is the value of interest (20 minutes),
- minutes (mean),
- minutes (standard deviation).
Now, using the standard normal distribution table or a calculator, the probability is approximately 0.0062.
Thus, the probability of waiting more than 20 minutes is 0.0062, or 0.62%.
Part (ii): Probability that the waiting time is less than 11 minutes
We use the same z-score formula for minutes:
From the standard normal distribution table, is approximately 0.0228.
Thus, the probability of waiting less than 11 minutes is 0.0228, or 2.28%.
Part (iii): Probability that the waiting time is between 10 and 18 minutes
We calculate the z-scores for both 10 and 18 minutes.
For minutes:
For minutes:
Using the standard normal table:
- is approximately 0.0062,
- is approximately 0.9332.
The probability of being between 10 and 18 minutes is the difference:
Thus, the probability is 0.9270, or 92.7%.
Would you like further details or any clarification on these steps?
Related Questions:
- What is the z-score for a waiting time of 12 minutes?
- How would the probabilities change if the standard deviation was 3 minutes?
- What is the 95th percentile of waiting times in this scenario?
- How do we calculate the probability for a range of values in a normal distribution?
- How does changing the mean impact the probabilities calculated above?
Tip: Z-scores help standardize values, making it easier to find probabilities in a normal distribution.
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Math Problem Analysis
Mathematical Concepts
Normal Distribution
Probability
Z-score
Formulas
Z = (x - μ) / σ
P(Z > z_value) = 1 - P(Z < z_value)
P(a < Z < b) = P(Z < b) - P(Z < a)
Theorems
Standard Normal Distribution
Suitable Grade Level
Grades 10-12
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