Question

The time required to travel downtown at 10 a.m. on Monday morning is known to be normally distributed with a mean of 40 minutes and a standard deviation of 5 minutes. What is the probability that it will take less than 40 minutes

Answer 1

Answer:

The probability that it will take less than 40 minutes is 0.50 or 50%.

Step-by-step explanation:

We are given that the time required to travel downtown at 10 a.m. on Monday morning is known to be normally distributed with a mean of 40 minutes and a standard deviation of 5 minutes.

Let X = time required to travel downtown at 10 a.m. on Monday morning

SO, X ~ N($\mu = 40,\sigma^{2} = 5^{2}$)

The z-score probability distribution is given by ;

                 Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = mean time = 40 minutes

            $\sigma$ = standard deviation = 5 minutes

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, the probability that it will take less than 40 minutes is given by = P(X < 40 minutes)

    P(X < 40) = P( $\frac{X-\mu}{\sigma}$ < $\frac{40-40}{5}$ ) = P(Z < 0) = 0.50

The above probability is calculated using z table by looking at value of x = 0 in the z table which have an area of 0.50 or 50%.

Therefore, probability that it will take less than 40 minutes 0.50 or 50%.