Answer:
a) P(X<100)=0.7855
P(X<200)=0.9418
P(100<X<200)=0.1830
b) P(X>M+2σ)=0.0498
c) Median = 48.74 m
Step-by-step explanation:
If the distance X is modeled by an exponential distribution, with parameter λ=0.01422, we have:
The cumulative probability function can be expressed as:

a) The probability that the distance is at most 100 m is:

The probability that the distance is at most 200 m is:

The probability that the distance is between 100 and 200 m is:

b) The mean is M=1/λ=70.3235 and the standard deviation σ=1/λ=70.3235.
So we have to calculate the probability that X exceeds 2 times the s.d. from the mean:

c) The value of the median distance is

3+4 = 7. 19/7 = 2.71
5+6 = 11. 41/11 = 3.73
1+3 = 4. ?/4 = 4.75 <--- Following the pattern of +1.02 (2.71+1.02=3.73, therefore 3.73+1.02=4.75
We still need (?) ... ?/4=4.75 we can re-write this as 4(4.75)=?
4(4.75) = 19 (4x4=16, 4x0.75=3, 16+3=19)
Therefore, 1+3=19.
Answear:
G=T hope dissssssssss helps
The answers are X=-4 and Y=3.