United Airlines' flights from Boston to Atlanta are on time 80 % of the time. Suppose 10 flights are randomly selected, and the number on-time flights is recorded.
Round answers to 3 significant figures.

The probability that exactly 8 flights are on time is =
0.301
Correct

The probability that at most 5 flights are on time is =

The probability that at least 7 flights are on time is =

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.

Binomial distribution

[tex]\large\boxed{X\sim \text{B}(n,p)}[/tex]

where:

X is the random variable that represents the number of successes.
n is the fixed number of independent trials.
p is the probability of success in each trial.

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:

[tex]n = 10[/tex]
[tex]p = 80\%=0.8[/tex]

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]

Part A

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]

Part B

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]

Part C

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).