Suppose a normal distribution has a mean of 79 and a standard deviation of 7. What is [tex]\( P(x \leq 65) \)[/tex]?

A. 0.84
B. 0.025
C. 0.975
D. 0.16



Answer :

Sure, let's address this step-by-step.

Firstly, we are given a normal distribution with:
- Mean ([tex]\(\mu\)[/tex]) = 79
- Standard deviation ([tex]\(\sigma\)[/tex]) = 7
- We seek the probability that a random variable [tex]\(X\)[/tex] is less than or equal to 65, i.e., [tex]\(P(X \leq 65)\)[/tex].

To find this probability, we need to calculate the z-score for [tex]\(X = 65\)[/tex]. The z-score tells us how many standard deviations a particular value [tex]\(X\)[/tex] is away from the mean. The formula for the z-score is given by:

[tex]\[ z = \frac{X - \mu}{\sigma} \][/tex]

Substituting the given values:

[tex]\[ z = \frac{65 - 79}{7} = \frac{-14}{7} = -2.0 \][/tex]

Next, we use the standard normal distribution to find the probability that corresponds to the z-score of [tex]\(-2.0\)[/tex]. This can be found using standard normal distribution tables or a cumulative distribution function (CDF) for the standard normal distribution.

For [tex]\(z = -2.0\)[/tex], the cumulative probability (area to the left of [tex]\(z\)[/tex]) is approximately [tex]\(0.02275\)[/tex].

Thus, the probability [tex]\(P(X \leq 65)\)[/tex] is 0.02275.

Now, let's match this value with the given options:
A. 0.84
B. 0.025
C. 0.975
D. 0.16

The closest value to 0.02275 provided in the options is 0.025.

So, the correct answer is:
B. 0.025