Suppose you have a binomial distribution with *n* = 29 and *p* = 0.4. Find P(6 ≤ X ≤ 7).

To solve the problem, we need to compute \( P(6 \leq X \leq 7) \) for a binomial distribution with \( n = 29 \) trials and probability \( p = 0.4 \).

The probability mass function (PMF) for a binomial distribution is given by:

\[
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
\]

Where:
- \( n = 29 \) (number of trials)
- \( p = 0.4 \) (probability of success in each trial)
- \( X \) is the number of successes

We want to find \( P(6 \leq X \leq 7) = P(X = 6) + P(X = 7) \).

Let's calculate both \( P(X = 6) \) and \( P(X = 7) \) using the PMF formula.

### Step-by-step:
\[
P(X = 6) = \binom{29}{6} (0.4)^6 (0.6)^{23}
\]
\[
P(X = 7) = \binom{29}{7} (0.4)^7 (0.6)^{22}
\]

I’ll calculate the exact values now.The probability \( P(6 \leq X \leq 7) \) for the given binomial distribution is approximately 0.0490.

Would you like more details on the steps or any clarifications?

Here are 5 related questions:
1. How would the probability change if \( p \) were 0.5 instead of 0.4?
2. Can you find the expected value and variance for this binomial distribution?
3. What is the cumulative probability \( P(X \leq 7) \) for this distribution?
4. How can this be approximated using the normal distribution?
5. How would you calculate \( P(X = 10) \) for this distribution?

**Tip**: In large binomial distributions, the normal approximation can simplify calculations for ranges of values.

Suppose you have a binomial distribution with n = 29 and p = 0.4. Find P(6 ≤ X ≤ 7).

Learn how to find P(6 ≤ X ≤ 7) for a binomial distribution with n = 29 and p = 0.4. This problem involves calculating probabilities using the binomial probability formula and applying the binomial theorem. Suitable for students in Grades 10-12.

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