Question

The time spent waiting in the line is approximately normally distributed. The mean waiting time is 6 minutes and the standard deviation of the waiting time is 2 minutes. Find the probability that a person will wait for more than 9 minutes. Round your answer to four decimal places.

Answer 1

Solution :

P(x > 9) = 1 - P(x < 9)

= 1 - P[(x - ) / < (9 - 6) / 2]  

= 1 - P(z < 1.5)

= 1 - 0.9332

= 0.0668

Probability = 0.0668

Answer 2

To find the probability that a person will wait for more than 9 minutes, we can use the standard normal distribution.

Step 1: Standardize the value 9 using the formula: z = (x - μ) / σ

where x is the given value (9 minutes), μ is the mean (6 minutes), and σ is the standard deviation (2 minutes).

Plugging in the values: z = (9 - 6) / 2 z = 3 / 2 z = 1.5

Step 2: Find the probability corresponding to the standardized value using a standard normal distribution table or a calculator.

Looking up the z-value of 1.5 in a standard normal distribution table, we find that the corresponding probability is 0.9332 (rounded to four decimal places).

Step 3: Subtract the probability from 1 to find the probability that a person will wait for more than 9 minutes: P(waiting time > 9 minutes) = 1 - 0.9332 P(waiting time > 9 minutes) ≈ 0.0668

Therefore, the probability that a person will wait for more than 9 minutes is approximately 0.0668, rounded to four decimal places.