You toss a fair coin 10000 times. what are the odds of obtaining more than 5100 tails, approximately?
1 answer:
This problem can be addressed by applying the normal approximation to a binomial distribution.
Calculations:
Mean (μ) = np = 10,000 × 0.5 = 5,000
The standard deviation (σ) is given by:
The probability of obtaining more than 5,100 tails is 0.0228, whereas the probability of fewer than 5,100 tails is 0.9772.
Thus, the odds of having more than 5,100 tails are:
0.0228 : 0.9772 = 1 : 42.86.
Calculations:
Mean (μ) = np = 10,000 × 0.5 = 5,000
The standard deviation (σ) is given by:
The probability of obtaining more than 5,100 tails is 0.0228, whereas the probability of fewer than 5,100 tails is 0.9772.
Thus, the odds of having more than 5,100 tails are:
0.0228 : 0.9772 = 1 : 42.86.
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Answer:Part a)
Part b) When Jenny divides the square root of her favorite positive integer by , the result is an integer.
Step-by-step explanation:
Let
x-------> the favorite positive integer
Part a)
1) For
-----> the product results in an integer
thus
The number could potentially be Jenny's favorite positive integer
2) For
-----> the product results in an integer
thus
The number could potentially be Jenny's favorite positive integer
3) For
-----> the product results in an integer
thus
The number could potentially be Jenny's favorite positive integer
Part B)
1) For
-----> the outcome is an integer
2) For
-----> the outcome is an integer
3) For
-----> the outcome is an integer
Therefore
When Jenny divides the square root of her favorite positive integer by , she obtains an integer as a result.