Question

A delivery company's trucks occasionally get parking tickets, and based on past experience, the company plans that the trucks will average 1.3 tickets a month, with a standard deviation of 0.7 tickets. If they have 18 trucks, what are the mean and standard deviation of the total number of parking tickets the company will have to pay this month?

Answer 1

If they have 18 trucks, what are the mean and standard deviation of the total number of parking tickets the company will have to pay this month?

• mean (18 trucks) = 23.4 tickets per month
• standard deviation (18 trucks) = 2.97 tickets

Step-by-step explanation:

since the mean of tickets per truck = 1.3 tickets, then for 18 trucks the mean = 1.3 tickets x 18 trucks = 23.4 tickets

variance = standard deviation² = 0.7² = 0.49

variance (18 trucks) = √(18 x 0.49) = √8.82 = 2.96985 ≈ 2.97 tickets

Answer 2

The mean of the tickets the company will pay this month is 23.4, and the standard deviation is 2.97

Step-by-step explanation:

Important data:

`\=x=1.3\\n=18\\s=0.7`

The mean of total number of parking tickets is:

`\mu_(x)=\=x.n\\\mu_(x)=(1.3)(18)= 23.4`

Standatrd deviation is given by:

`\sigma_(x)=\sqrt{s^(2)n } \\\sigma_(x)=\sqrt{(0.7)^(2)(18) } \\\sigma_(x)=√(8.82)\\ \sigma_(x)=2.97`

The mean of the tickets the company will pay this month is 23.4, and the standard deviation is 2.97