In 100 trials of this experiment, it is expected that about 7 will result in exactly 11 flights being on time. (Round to the nearest whole number as needed.) (d) Using the binomial distribution, the probability that fewer than 11 flights are on time is 0.0377 (Round to four decimal places as needed.) Interpret the probability. In 100 trials of this experiment, it is expected that about 4 will result in fewer than 11 flights being on time. (Round to the nearest whole number as needed.) (e) Using the binomial distribution, the probability that at least 11 flights are on time is 0.9623 . (Round to four decimal places as needed.) Interpret the probability. In 100 trials of this experiment, it is expected that about 96 will result in at least 11 flights being on time. (Round to the nearest whole number as needed.) (f) Using the binomial distribution, the probability that between 9 and 11 flights, inclusive, are on time is (Round to four decimal places as needed.) (Rol.

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To solve the problem, we will break it down into parts (d), (e), and (f) as specified.

### Part (d)
**Given:**
- Probability that fewer than 11 flights are on time: $$ P(X < 11) = 0.0377 $$

**Interpretation:**
This probability means that in a single trial of the experiment, there is a 3.77% chance that fewer than 11 flights will be on time.

**Expected number of trials:**
In 100 trials, the expected number of trials resulting in fewer than 11 flights being on time is calculated as:
$$
\text{Expected trials} = 100 \times P(X < 11) = 100 \times 0.0377
$$
Calculating this gives:
$$
\text{Expected trials} = 3.77
$$
Rounding to the nearest whole number, we get:
$$
\text{Expected trials} \approx 4
$$

### Part (e)
**Given:**
- Probability that at least 11 flights are on time: $$ P(X \geq 11) = 0.9623 $$

**Interpretation:**
This probability means that in a single trial of the experiment, there is a 96.23% chance that at least 11 flights will be on time.

**Expected number of trials:**
In 100 trials, the expected number of trials resulting in at least 11 flights being on time is calculated as:
$$
\text{Expected trials} = 100 \times P(X \geq 11) = 100 \times 0.9623
$$
Calculating this gives:
$$
\text{Expected trials} = 96.23
$$
Rounding to the nearest whole number, we get:
$$
\text{Expected trials} \approx 96
$$

### Part (f)
**To find the probability that between 9 and 11 flights, inclusive, are on time:**
This can be expressed as:
$$
P(9 \leq X \leq 11) = P(X = 9) + P(X = 10) + P(X = 11)
$$

We will need to calculate these probabilities using the binomial distribution formula:
$$
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
$$
where:
- $$ n = 100 $$ (number of trials)
- $$ p $$ is the probability of a flight being on time (not provided, but we can assume it is derived from the context)
- $$ k $$ is the number of successes (9, 10, or 11)

Since the probability $$ p $$ is not given, we will assume it is derived from the context of the problem. However, if you have a specific value for $$ p $$, please provide it, and I can calculate the probabilities for $$ P(X = 9) $$, $$ P(X = 10) $$, and $$ P(X = 11) $$.

If you want to proceed with a specific value for $$ p $$, please provide it, or I can proceed with a general calculation assuming a hypothetical value.
**Summary of Probabilities:** - **Part (d):** In 100 trials, approximately 4 trials are expected to have fewer than 11 flights on time. - **Part (e):** In 100 trials, approximately 96 trials are expected to have at least 11 flights on time. - **Part (f):** The probability that between 9 and 11 flights are on time is calculated using the binomial distribution formula. To find the exact probability, the probability $$ p $$ of a single flight being on time is needed.

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