1 answer

Bus trip times between 2 stops on a certain route are normally distributed with a mean...

Question:

Bus trip times between 2 stops on a certain route are normally distributed with a mean of 5.7 minutes and a standard deviatio
Bus trip times between 2 stops on a certain route are normally distributed with a mean of 5.7 minutes and a standard deviation of 1.5 minutes. What is the trip time that constitutes the 35th percentile? Express your answer correct to 1 decimal place and add the units (minutes) /content/enforced/490502-MATH210002_2020W/z-table-1.pdf

Answers

Answer is:

{\color{DarkRed} 6.3 \text{ minutes}}

Calculations

We know that X is normally distributed, with parameters:

\mu = 5.7, \text{ } \sigma = 1.5

We need to find a score x so that the corresponding cumulative normal probability is equal to 0.35. Mathematically, x is such that:

\Pr(X \le x) = 0.35

The corresponding z score so that the cumulative standard normal probability distribution is 0.35 is

z_c = 0.3853

This value of zc​=0.3853 can be found either with Excel, or with a normal distribution table. Hence, the X score associated with the 0.35 cumulative probability is

x = \mu + z_c \times \sigma = 5.7 + 0.3853 \times 1.5 = 6.278 \sim 6.3 \text{ minutes}

Graphically:

normaldistributiongrapher.php?mean=5.7&s

Let me know in the comments if anything is not clear. I will reply ASAP! Please do upvote if satisfied!

.

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