To determine the probability [tex]\(P(X < 13)\)[/tex] for a normally distributed random variable [tex]\(X\)[/tex] with a mean ([tex]\(\mu\)[/tex]) of 10 and a standard deviation ([tex]\(\sigma\)[/tex]) of 2, we'll follow these steps:
1. Identify the problem:
- Given: [tex]\(\mu = 10\)[/tex], [tex]\(\sigma = 2\)[/tex]
- Find: [tex]\(P(X < 13)\)[/tex]
2. Standardizing [tex]\(X\)[/tex]:
- To find [tex]\(P(X < 13)\)[/tex], we standardize [tex]\(X\)[/tex] by converting it to a standard normal variable [tex]\(Z\)[/tex], which has a mean of 0 and a standard deviation of 1.
- The standardized variable [tex]\(Z\)[/tex] can be calculated using the formula:
[tex]\[
Z = \frac{X - \mu}{\sigma}
\][/tex]
- Here, [tex]\(X = 13\)[/tex], so:
[tex]\[
Z = \frac{13 - 10}{2} = \frac{3}{2} = 1.5
\][/tex]
3. Using the cumulative distribution function (CDF):
- The probability [tex]\(P(X < 13)\)[/tex] is equivalent to [tex]\(P(Z < 1.5)\)[/tex], where [tex]\(Z\)[/tex] follows the standard normal distribution.
- We use the cumulative distribution function (CDF) of the standard normal distribution to find [tex]\(P(Z < 1.5)\)[/tex].
4. Consult a standard normal distribution table or use a CDF calculator:
- Looking up the value for [tex]\(Z = 1.5\)[/tex] in the standard normal distribution table or using a CDF calculator, we get the probability.
5. Result:
- From the standard normal distribution table or calculator, we find that [tex]\(P(Z < 1.5) \approx 0.93319\)[/tex].
Therefore, the probability [tex]\(P(X < 13)\)[/tex] for a normally distributed random variable [tex]\(X\)[/tex] with mean 10 and standard deviation 2 is approximately [tex]\(0.93319\)[/tex].
