Question

the time taken to clean up a particular fast food outlet after it closes follows a normal distribution with a mean of 30 minutes and a standard deviation off 5minutes.

a. what is the probability that the cleanup crew will complete the job in less than 20 minutes.

b. what is the probability that the cleanup crew will complete the job in more than 38 minutes.

Answer 1

Solution :

Given that ,

mean = $\mu$ = 30

standard deviation = $\sigma$ = 5

a.

P(x < 20) = P[(x - $\mu$ ) / $\sigma$ < (20 - 30) / 5]

= P(z < -2)

= 0.0228

Probability = 0.0228

b.

P(x > 38) = 1 - P(x < 38)

= 1 - P[(x - $\mu$ ) / $\sigma$ < (38 - 30) / 5]

= 1 - P(z < 1.6)

= 1 - 0.9452  

= 0.0548

Probability = 0.0548