Using the information given, it is found that the class width for this frequency distribution table is of 1.
In this problem, these following classes are given:
0 – 1 14
2 – 3 1
4 – 5 8
6 – 7 12
8 – 9 12
The classes not given, which are 1 - 2, 3 - 4 and 5 - 6, have values of 0.
The <u>difference between the bounds of the classes is of 1</u>, thus, the class width is of 1.
A similar problem is given at brainly.com/question/24701109
Answer: 0.6767
Step-by-step explanation:
Given : Mean =
errors per page
Let X be the number of errors in a particular page.
The formula to calculate the Poisson distribution is given by :_
Now, the probability that a randomly selected page does not need to be retyped is given by :-
Hence, the required probability :- 0.6767
Answer:
They both have 1/2
Step-by-step explanation:
For the first one there are 6 numbers and 3 of them are odd. That makes it 1/2.
For the second one it is self explanatory.
Dawson's annual premium will be $2,462.40. This can be found by going across from "Male 40-44" over to "20-year coverage" which is $13.68. Since $13.68 is per $1000 of coverage, you would multiply it by 180 to get $2,462.40.
Rachel's checks I believe would have a deduction of $63.14.
