Math Problem Statement

solve all parts completely

In a supermarket, the time customers spend waiting for their turn at the cashier follows a normal distribution with a mean of 15 minutes and standard deviation 2 minutes. Find the probability that the waiting time is (i) More than 20 minutes, (ii) Less than 11 minutes, (iii) Between 10 and 18 minutes.

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:

z=xμσz = \frac{x - \mu}{\sigma}

Where:

  • xx is the value of interest (20 minutes),
  • μ=15\mu = 15 minutes (mean),
  • σ=2\sigma = 2 minutes (standard deviation).

z=20152=52=2.5z = \frac{20 - 15}{2} = \frac{5}{2} = 2.5

Now, using the standard normal distribution table or a calculator, the probability P(Z>2.5)P(Z > 2.5) 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 x=11x = 11 minutes:

z=11152=42=2z = \frac{11 - 15}{2} = \frac{-4}{2} = -2

From the standard normal distribution table, P(Z<2)P(Z < -2) 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 x=10x = 10 minutes:

z=10152=2.5z = \frac{10 - 15}{2} = -2.5

For x=18x = 18 minutes:

z=18152=1.5z = \frac{18 - 15}{2} = 1.5

Using the standard normal table:

  • P(Z<2.5)P(Z < -2.5) is approximately 0.0062,
  • P(Z<1.5)P(Z < 1.5) is approximately 0.9332.

The probability of being between 10 and 18 minutes is the difference:

P(10<X<18)=P(Z<1.5)P(Z<2.5)=0.93320.0062=0.9270P(10 < X < 18) = P(Z < 1.5) - P(Z < -2.5) = 0.9332 - 0.0062 = 0.9270

Thus, the probability is 0.9270, or 92.7%.


Would you like further details or any clarification on these steps?

Related Questions:

  1. What is the z-score for a waiting time of 12 minutes?
  2. How would the probabilities change if the standard deviation was 3 minutes?
  3. What is the 95th percentile of waiting times in this scenario?
  4. How do we calculate the probability for a range of values in a normal distribution?
  5. 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