To calculate [tex]\( P(X \leq 49) \)[/tex] for a binomially distributed random variable [tex]\( X \)[/tex] with [tex]\( n = 100 \)[/tex] and [tex]\( p = 0.6 \)[/tex] using the normal approximation to the binomial distribution, follow these steps:
1. Determine the mean and standard deviation for the binomial distribution:
The mean [tex]\( \mu \)[/tex] of a binomial distribution is given by:
[tex]\[
\mu = n \cdot p
\][/tex]
For [tex]\( n = 100 \)[/tex] and [tex]\( p = 0.6 \)[/tex], we have:
[tex]\[
\mu = 100 \cdot 0.6 = 60
\][/tex]
The standard deviation [tex]\( \sigma \)[/tex] of a binomial distribution is given by:
[tex]\[
\sigma = \sqrt{n \cdot p \cdot (1 - p)}
\][/tex]
So, we have:
[tex]\[
\sigma = \sqrt{100 \cdot 0.6 \cdot 0.4} = \sqrt{24} \approx 4.899
\][/tex]
2. Apply the continuity correction and convert to the standard normal distribution:
Since we are using the normal approximation, we need to apply the continuity correction by adding 0.5 to the value we are assessing:
[tex]\[
P(X \leq 49) \approx P\left(Y \leq 49.5\right)
\][/tex]
where [tex]\( Y \)[/tex] is the normal approximation of the binomial distribution.
Convert this to the standard normal distribution (Z-score):
[tex]\[
Z = \frac{49.5 - \mu}{\sigma}
\][/tex]
Plugging in the values:
[tex]\[
Z = \frac{49.5 - 60}{4.899} \approx \frac{-10.5}{4.899} \approx -2.143
\][/tex]
3. Find the cumulative probability associated with the Z-score:
Next, use the cumulative distribution function (CDF) of the standard normal distribution to find the probability corresponding to the Z-score:
[tex]\[
P(Z \leq -2.143) \approx 0.016
\][/tex]
Therefore, [tex]\( P(X \leq 49) \)[/tex] is approximately [tex]\( 0.016 \)[/tex].
