Answer :
Answer:
The mean of the tickets the company will pay this month is 23.4, and the standard deviation is 2.97
Explanation:
Important data:
[tex]\=x=1.3\\n=18\\s=0.7[/tex]
The mean of total number of parking tickets is:
[tex]\mu_{x}=\=x.n\\\mu_{x}=(1.3)(18)= 23.4[/tex]
Standatrd deviation is given by:
[tex]\sigma_{x}=\sqrt{s^{2}n } \\\sigma_{x}=\sqrt{(0.7)^{2}(18) } \\\sigma_{x}=\sqrt{8.82}\\ \sigma_{x}=2.97[/tex]
The mean of the tickets the company will pay this month is 23.4, and the standard deviation is 2.97
Answer:
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
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