For \( n=10 \) and \( p=0.9 \), check the binomial distribution to see whether it can be approximated by a normal distribution.

1. **Identify the parameters of the binomial distribution.** The parameters are: - Number of trials: \( n = 10 \) - Probability of success: \( p = 0.9 \) 2. **Compute the mean and variance.** - The mean is \[ \mu = np = 10 \times 0.9 = 9. \] - The variance is \[ \sigma^2 = np(1-p) = 10 \times 0.9 \times 0.1 = 0.9. \] - The standard deviation is \[ \sigma = \sqrt{0.9} \approx 0.95. \] 3. **Check the usual conditions for normal approximation of a binomial distribution.** A common rule of thumb for using the normal approximation is that: - \( np \) should be sufficiently large (often \( \geq 5 \) or \( \geq 10 \)), and - \( n(1-p) \) should also be sufficiently large. Here: - \( np = 9 \) (which is barely acceptable in some contexts), and - \( n(1-p) = 10 \times 0.1 = 1 \) (which is too small). 4. **Conclusion.** Since \( n(1-p) = 1 \) is much less than the typical threshold (commonly 5 or 10), the binomial distribution for \( n = 10 \) and \( p = 0.9 \) is **not** well approximated by a normal distribution.