Certainly! Let's go through the steps to find the probability [tex]\( P(x > 36) \)[/tex] for a normally distributed random variable [tex]\( x \)[/tex] with a mean [tex]\( \mu = 50 \)[/tex] and standard deviation [tex]\( \sigma = 7 \)[/tex].
1. Standardize the Random Variable:
To find the probability, we first need to convert the random variable [tex]\( x \)[/tex] into its corresponding z-score. The z-score is given by:
[tex]\[
z = \frac{x - \mu}{\sigma}
\][/tex]
Here, [tex]\( x = 36 \)[/tex], [tex]\( \mu = 50 \)[/tex], and [tex]\( \sigma = 7 \)[/tex]. Plugging in these values:
[tex]\[
z = \frac{36 - 50}{7} = \frac{-14}{7} = -2
\][/tex]
So, the z-score corresponding to [tex]\( x = 36 \)[/tex] is [tex]\( z = -2 \)[/tex].
2. Find the Cumulative Probability:
To find the probability related to this z-score, we use the cumulative distribution function (CDF) of the standard normal distribution. The CDF [tex]\( \Phi(z) \)[/tex] gives us the probability that a standard normal variable is less than or equal to a given z-score:
[tex]\[
P(z \leq -2)
\][/tex]
The cumulative probability for [tex]\( z \leq -2 \)[/tex] is [tex]\( \Phi(-2) \)[/tex], which is approximately 0.02275.
3. Calculate the Desired Probability:
We need the probability that [tex]\( x > 36 \)[/tex]. This is the complement of the cumulative probability we just found:
[tex]\[
P(x > 36) = 1 - P(x \leq 36)
\][/tex]
Since [tex]\( P(x \leq 36) \)[/tex] is the same as [tex]\( \Phi(-2) \)[/tex]:
[tex]\[
P(x > 36) = 1 - \Phi(-2) = 1 - 0.02275 = 0.97725
\][/tex]
Therefore, the probability that [tex]\( x \)[/tex] exceeds 36 is approximately [tex]\( 0.97725 \)[/tex].
So, [tex]\( P(x > 36) \)[/tex] is approximately 0.97725, or in percentage terms, approximately 97.725%.
For clarity, here are the key results:
- The z-score for [tex]\( x = 36 \)[/tex] is -2.
- [tex]\( P(x \leq 36) = 0.02275 \)[/tex]
- [tex]\( P(x > 36) = 0.97725 \)[/tex].
