John kept track of how many baskets were made in a basketball game. After 4 minutes, 5 baskets were made. How long did it take u
2 answers:
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Answer:
The standard deviation for the mean weigth of Salmon is 2/3 lbs for restaurants, 2/7 lbs for grocery stores and 1/4 lbs for discount order stores.
Step-by-step explanation:
The mean sample of the sum of n random variables is
If are indentically distributed and independent, like in the situation of the problem, then the variance of
will be the sum of the variances, in other words, it will be n times the variance of
.
However if we multiply this mean by 1/n (in other words, divide by n), then we have to divide the variance by 1/n², thus and as a result, the standard deviation of
is the standard deviation of
divided by
.
Since the standard deviation of the weigth of a Salmon is 2 lbs, then the standard deviations for the mean weigth will be:
- Restaurants: We have boxes with 9 salmon each, so it will be
- Grocery stores: Each carton has 49 salmon, thus the standard deviation is
- Discount outlet stores: Each pallet has 64 salmon, as a result, the standard deviation is
We conclude that de standard deivation of the mean weigth of salmon of the types of shipment given is: 2/3 lbs for restaurants, 2/7 lbs for grocery stores and 1/4 lbs for discount outlet stores.
Answer:
b) No, it's not independent.
c) 0.02
d) 0.59
e) 0.57
f) 0.5616
Step-by-step explanation:
To answer this problem, a Venn diagram should be useful. The diagram with the information of Event 1 and Event 2 is shown below (I already added the information for the intersection but we're going to see how to get that information in the b) part of the problem)
Let's call A the event that she passes the first course, then P(A)=.73
Let's call B the event that she passes the second course, then P(B)=.66
Then P(A∪B) is the probability that she passes the first or the second course (at least one of them) is the given probability. P(A∪B)=.98
b) Is the event she passes one course independent of the event that she passes the other course?
Two events are independent when P(A∩B) = P(A) * P(B)
So far, we don't know P(A∩B), but we do know that for all events, the next formula is true:
P(A∪B) = P(A) + P(B) - P(A∩B)
We are going to solve for P (A∩B)
.98 = .73 + .66 - P(A∩B)
P(A∩B) =.73 + .66 - .98
P(A∩B) = .41
Now we will see if the formula for independent events is true
P(A∩B) = P(A) x P(B)
.41 = .73 x .66
.41 ≠.4818
Therefore, these two events are not independent.
c) The probability she does not pass either course, is 1 - the probability that she passes either one of the courses (P(A∪B) = .98)
1 - P(A∪B) = 1 - .98 = .02
d) The probability she doesn't pass both courses is 1 - the probability that she passes both of the courses P(A∩B)
1 - P(A∩B) = 1 -.41 = .59
e) The probability she passes exactly one course would be the probability that she passes either course minus the probability that she passes both courses.
P(A∪B) - P(A∩B) = .98 - .41 = .57
f) Given that she passes the first course, the probability she passes the second would be a conditional probability P(B|A)
P(B|A) = P(A∩B) / P(A)
P(B|A) = .41 / .73 = .5616
