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Answer:
a) The probability that exactly 8 flights are on time = 0.302
b) The probability that at most 5 flights are on time is = 0.0328
c) The probability that at least 7 flights are on time is = 0.879
Step-by-step explanation:
Please find the attached.
Answer:
The probability that exactly 8 flights are on time is = 0.302
The probability that at most 5 flights are on time is = 0.0328
The probability that at least 7 flights are on time is = 0.879
Step-by-step explanation:
We can model the given scenario as a binomial distribution, since we are dealing with a situation where flights are either on time or not on time (success or failure), and the probability of success (flight on time) remains constant.
[tex]\large\boxed{X\sim \text{B}(n,p)}[/tex]
where:
Since we want to find the probability of the flights being on time, we consider "on time" as the success and "not on time" as the failure, so we have:
Therefore:
[tex]\large\boxed{X\sim \text{B}(10,0.8)}[/tex]
where the random variable X represents the number of flights that are on time.
[tex]\dotfill[/tex]
To find the probability that exactly 8 flights are on time, we need to find P(X = 8).
To do this, we can either use the binomial probability formula, or enter the values into a statistical calculator.
The binomial probability formula is:
[tex]\displaystyle P(X=x)=\binom{n}{x} \cdot p^x \cdot (1-p)^{n-x}[/tex]
In this case, n = 10, x = 8, and p = 0.8.
Therefore:
[tex]\displaystyle P(X = 8) = \binom{10}{8} \times 0.8^{8} \times (1 - 0.8)^{10 - 8}\\\\ P(X = 8) = 45 \times 0.8^{8} \times 0.2^{2}\\\\ P(X = 8) = 45 \times 0.16777216 \times 0.04\\\\ P(X = 8) = 0.301989888\\\\P(X = 8) = 0.302 \;\sf(3\;s.f.)[/tex]
So, the probability that exactly 8 flights are on time is 0.301 (rounded to three significant figures).
[tex]\dotfill[/tex]
To find the probability that at most 5 flights are on time, we need to find P(X ≤ 5).
To do this, we can use the binomial cumulative distribution function on a calculator. As the calculator function gives us P(X ≤ x) for X ~ B(n, p), we enter x = 5, n = 10, p - 0.8:
[tex]P(X\leq 5)=0.0327934975... \\\\P(X\leq 5)=0.0328\; \sf (3\;s.f.)[/tex]
So, the probability that at most 5 flights are on time is 0.0328 (rounded to three significant figures).
[tex]\dotfill[/tex]
To find the probability that at least 7 flights are on time, we need to find P(X ≥ 7).
To do this, we can use the binomial cumulative distribution function on a calculator. As the calculator function gives us P(X ≤ x) for X ~ B(n, p), we will need to calculate P(X ≤ 6) and then subtract it from 1:
[tex]P(X \geq 7)=1-P(X \leq 6) \\\\P(X \geq 7)=1-0.120873881... \\\\P(X \geq 7)=0.879126118...\\\\P(X \geq 7)=0.879\; \sf (3\;s.f.)[/tex]
So, the probability that at least 7 flights are on time is 0.879 (rounded to three significant figures).