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Answer:
Probability of N Divisible by 3 - 0.33
Probability of N Divisible by 5 - 0.2
Probability of N Divisible by 7 - 0.413
Probability of N Divisible by 15 - 0.066
Probability of N Divisible by 105 - 0.0095
Step-by-step explanation:
Given data:
Integer N {1,2,.....10^3}
Thus total number of ways by which 1000 is divisible by 3 i.e. 1000/3 = 333.3
Probability of N divisible by 3 {N%3 = 0 }
total number of ways by which 1000 is divisible by 5 i.e. 1000/5 = 200
Probability of N divisible by 5 {N%5 = 0 }
total number of ways by which 1000 is divisible by 7 i.e. 1000/7 = 142.857
Probability of N divisible by 7 {N%7 = 0 }
total number of ways by which 1000 is divisible by 15 i.e. 1000/15 = 66.667
Probability of N divisible by 15 {N%15 = 0 }
total number of ways by which 1000 is divisible by 105 i.e. 1000/105 = 9.52
Probability of N divisible by 105 {N%105 = 0 }
similarly for N is selected from 1,2.....(10)^k where K is large then the N value. Therefore effect of k will remain same as previous part.
Answer:
a)
b)
c) We want to find a value c who satisfy this condition:
And using the cumulative distribution function we have this:
And solving for c we got:
Step-by-step explanation:
For this case we define the random variable X as he amount of time (in minutes) that a particular San Francisco commuter must wait for a BART train, and we know that the distribution for X is given by:
Part a
We want this probability:
And for this case we can use the cumulative distribution function given by:
And using the cumulative distribution function we got:
For the probability if we use the cumulative distribution function and the complement rule we got:
Part b
We want this probability:
And using the cdf we got:
Part c
We want to find a value c who satisfy this condition:
And using the cumulative distribution function we have this:
And solving for c we got: