Given Information:
Probability = p = 0.2
Number of trials = n = 21 days
Required Information:
(a) P(x ≥ 3) = ?
(b) P(x = 5) = ?
c) is the sampling distribution of the sample proportion approximately normal?
Answer:
(a) P(x ≥ 3) = 0.8213
(b) P(x = 5) = 0.1829
(c) we can conclude that the sampling distribution of the sample proportion is not approximately normal because the condition np > 5 is not satisfied.
Step-by-step explanation:
We know that a binomial distribution is given by
P(x) = ⁿCₓ pˣ (1 - p)ⁿ⁻ˣ
Where n is the number of trials, x is the variable of interest and p is the probability of success and (1 - p) is the probability of failure.
(a) For the next 21 days that Sasha travels to work, what is the probability that Sasha will experience a delay due to traffic on at least 3 of the days?
At least 3 means equal or greater than 3
P(x ≥ 3) = 1 - P(x < 3)
P(x ≥ 3) = 1 - [P(x = 0) + P(x = 1) + P(x = 2)]
For P(x = 0):
Here we have x = 0, n = 21 and p = 0.2
P(x = 0) = ²¹C₀*0.2⁰*(1 - 0.2)²¹⁻⁰
P(x = 0) = 1*0.2⁰*(0.8)²¹
P(x = 0) = 0.00922
For P(x = 1):
Here we have x = 1, n = 21 and p = 0.2
P(x = 1) = ²¹C₁*0.2¹*(1 - 0.2)²¹⁻¹
P(x = 1) = 21*0.2¹*(0.8)²⁰
P(x = 1) = 0.0484
For P(x = 2):
Here we have x = 2, n = 21 and p = 0.2
P(x = 2) = ²¹C₂*0.2²*(1 - 0.2)²¹⁻²
P(x = 2) = 210*0.2²*(0.8)¹⁹
P(x = 2) = 0.12105
Finally,
P(x ≥ 3) = 1 - [P(x = 0) + P(x = 1) + P(x = 2)]
P(x ≥ 3) = 1 - [0.00922 + 0.0484 + 0.12105]
P(x ≥ 3) = 1 - [0.17867]
P(x ≥ 3) = 0.8213
(b) What is the probability that Sasha’s first delay due to traffic will occur after the fifth day of travel to work?
Delay occur after the fifth day so that means for the first four days delay doesn't occur
P(x = 5) = 0.2¹*(1 - 0.2)⁵⁻¹
P(x = 5) = 0.2¹*(0.8)⁴
P(x = 5) = 0.1829
(c) Consider a random sample of 21 days that Sasha will travel to work. For the proportion of those days that she will experience a delay due to traffic, is the sampling distribution of the sample proportion approximately normal?
If both of these conditions are satisfied then the sampling distribution of the sample proportion is approximately normal.
np > 5 and nq > 5
Here n is the number of trials that is 21 and p is the probability of success that is 0.2 and q is the probability of failure that is 0.8.
nq > 5
21*0.8 > 5
16.8 > 5 ( satisfied)
np > 5
21*0.2 > 5
4.2 > 5 (not satisfied)
Therefore, we can conclude that the sampling distribution of the sample proportion is not approximately normal.
