Marvin lives in Stormwind City and works as an engineer in the city of Ironforge. In the morning, he has 333 transportation opti
1 answer:
The event "Atleast once" is the complement of event "None".
So, the probability that Marvin teleports atleast once per day will the compliment of probability that he does not teleports during the day. Therefore, first we need to find the probability that Marvin does not teleports during the day.
At Morning, the probability that Marvin does not teleport = 2/3
Likewise, the probability tha Marvin does not teleport during evening is also 2/3.
Since the two events are independent i.e. his choice during morning is not affecting his choice during the evening, the probability that he does not teleports during the day will be the product of both individual probabilities.
So, the probability that Marvin does not teleport during the day =
Probability that Marvin teleports atleast once during the day = 1 - Probability that Marvin does not teleport during the day.
Probability that Marvin teleports atleast once during the day =
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Answer:
a) Data given: 42 36 48 51 39 39 42 36 48 33 39 42 45
b) 44.1, 37.8, 50.4, 53.55, 40.95, 44.1, 37.8, 50.4 ,34.65, 40.95, 44.1, 47.25
c) 3.5, 3, 4, 4.25, 3.25, 3.25, 3.5, 3, 4, 2.75, 3.25, 3.5, 3.75
d) As we can see, the average of part b is 1.05 times the average of part a (1.05 * 41.538 = 43.615) and the average of part c is equal to the average obtained in part a divided by 12 (41.538 / 12 = 3.462).
And that happens because we create linear transformations for the parts b and c and the linear transformation affects the mean.
And you have the same interpretation for the deviation, it is affected by the linear transformation as the mean.
Step-by-step explanation:
For this case we can use the following formulas for the mean and standard deviation:
Part a
Data given: 42 36 48 51 39 39 42 36 48 33 39 42 45
And if we calculate the mean we got:
Part b
For this case we know that each value present a 5% of rise so we just need to multiply each value bu 1.05 and we have this new dataset:
44.1, 37.8, 50.4, 53.55, 40.95, 44.1, 37.8, 50.4 ,34.65, 40.95, 44.1, 47.25
And if we calculate the new mean and deviation we got:
Part c
The new dataset would be each value divided by 12 so we have:
3.5, 3, 4, 4.25, 3.25, 3.25, 3.5, 3, 4, 2.75, 3.25, 3.5, 3.75
And the new mean and deviation would be:
Part d
As we can see, the average of part b is 1.05 times the average of part a (1.05 * 41.538 = 43.615) and the average of part c is equal to the average obtained in part a divided by 12 (41.538 / 12 = 3,462).
And that happens because we create linear transformations for the parts b and c and the linear transformation affects the mean.
And you have the same interpretation for the deviation, it is affected by the linear transformation as the mean.