Question

The amount of time workers spend commuting to their jobs each day in a large metropolitan city has a mean of 70 minutes and a standard deviation of 20 minutes. Assuming the distribution of commuting times is known to be mound-shaped and symmetric, what percentage of these commuting times are between 50 and 110 minutes?

Answer 1

Answer:

81.85% of the workers spend between 50 and 110 commuting to work

Step-by-step explanation:

We can assume that the distribution is Normal (or approximately Normal) because we know that it is symmetric and mound-shaped.

We call X the time spend from one worker; X has distribution N(μ = 70, σ = 20). In order to make computations, we take W, the standarization of X, whose distribution is N(0,1)

$W = \frac{X-μ}{σ} = \frac{X-70}{20}$

The values of the cummulative distribution function of the standard normal, which we denote $\phi$ , are tabulated. You can find those values in the attached file.

$P(50 < X < 110) = P(\frac{50-70}{20} < \frac{X-70}{20} < \frac{110-70}{20}) = P(-1 < W < 2) = \\\phi(2) - \phi(-1)$

Using the symmetry of the Normal density function, we have that $\phi(-1) = 1-\phi(1)$ . Hece,

$P(50 < X < 110) = \phi(2) - \phi(-1) = \phi(2) - (1-\phi(1)) = \phi(2) + \phi(1) - 1 = \\0.9772+0.8413-1 = 0.8185$

The probability for a worker to spend that time commuting is 0.8185. We conclude that 81.85% of the workers spend between 50 and 110 commuting to work.

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