If Y gas a geometric distribution with success probability 0.3,what is the largest value of y such that P(Y>y)≥0.1
I had my answer to be 6.But I'm not sure
P=0.3
q=1-p =0.7
P(Y=y)=q^y=0.7^y
y such that 0.7^y=0.1
I apply ln to both sides and I get y to be 6
But I'm not sure with my answer
To find the largest value of y such that P(Y > y) ≥ 0.1, we can use the formula for the cumulative distribution function (CDF) of a geometric distribution:
CDF(y) = 1 - (1 - p)^y
where p is the success probability.
We want to find the largest value of y for which CDF(y) ≥ 0.1.
Plug in p = 0.3:
CDF(y) = 1 - (1 - 0.3)^y
Now, let's evaluate CDF(y) for various values of y to find the largest value that satisfies CDF(y) ≥ 0.1:
CDF(0) = 1 - (0.7)^0 = 1
CDF(1) = 1 - (0.7)^1 ≈ 0.7
CDF(2) = 1 - (0.7)^2 ≈ 0.91
CDF(3) = 1 - (0.7)^3 ≈ 0.973
CDF(4) = 1 - (0.7)^4 ≈ 0.991
CDF(5) = 1 - (0.7)^5 ≈ 0.997
CDF(6) = 1 - (0.7)^6 ≈ 0.999
As you can see, the largest value of y for which CDF(y) ≥ 0.1 is 6. Therefore, your answer of 6 is correct.
To find the largest value of y such that P(Y > y) ≥ 0.1, we can use the geometric distribution formula:
P(Y > y) = (1 - p)^y
Where:
P(Y > y) represents the probability that Y is greater than y.
p is the success probability.
In this case, p = 0.3 (success probability).
Let's substitute the values into the formula:
P(Y > y) = (1 - 0.3)^y
Now, we need to find the largest value of y such that P(Y > y) ≥ 0.1:
(1 - 0.3)^y ≥ 0.1
0.7^y ≥ 0.1
To solve for y, we can take the logarithm of both sides:
log(0.7^y) ≥ log(0.1)
Using the property of logarithms, we can bring down the exponent:
y * log(0.7) ≥ log(0.1)
Divide both sides by log(0.7):
y ≥ log(0.1) / log(0.7)
Calculating the right-hand side:
y ≥ -1 / 0.1549
y ≥ -6.46
Since we are looking for the largest integer value of y, y = -6 is not a valid solution. Therefore, the largest value of y such that P(Y > y) ≥ 0.1 is y = -5.