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

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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.
The binomial distribution with $$ n = 10 $$ and $$ p = 0.9 $$ **cannot** be approximated by a normal distribution because $$ n(1-p) = 1 $$ is too small.

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

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