Statistics- Finding Probability

Question

A local lawn service has determined the average time it takes to mow an average residential yard is thirty-five minutes. If mowing times are independent and constant, what is the probability it will take no more than forty-five minutes to mow a given yard?

It DOES NOT give the standard deviation. Am I supposed to assume a certain standard deviation because the term "independent and constant" connotes a certain probability distribution function? With the standard deviation, we could find the $z$-score and find $p$ that way. Otherwise, is there a different way to solve this problem?

Answer 1

Four speculative scenarios follow: Perhaps the context of the problem in the text or lectures could provide a clue which (if any) might be intended.

1) Uniform. Perhaps, somehow, "constant" is supposed to suggest a uniform distribution. To have mean 35, that would have to be $X \sim Unif(0, 70),$ so $P(X \le 45) = 45/70 = 0.6429.$

2) Normal. Some yards may take almost no time and some much more. If a normal distribution is assumed that might imply $\sigma \approx 10$ so that zero is a little more than three SD below the mean. Then $X \sim Norm(\mu=35, \sigma=10),$ and $P(X \le 45) \approx .84.$ from the 'Empirical Rule' or normal tables. [As suggested in one Comment by @AndreNicholas, but with the Empirical Rule instead of Markov Inequality.]

3) Exponential. Often when $\mu$ is given and $\sigma$ is not, then one is expected to assume an exponential distribution (for which $\mu = \sigma).$ In that case, $P(X \le 45) = 7235.$ [As suggested in the comment by @Blaine.]

4) Degenerate. If 'constant' is taken to be 'degenerate' (at 35), then it's a trick questions with $P(X \le 45) = 1.$

Computations from R:

 punif(45, 0, 70)
 ## 0.6428571
 pnorm(45, 35, 10)
 ## 0.8413447
 pexp(45, 1/35)
 ## 0.723547