Question

suppose a normal distribution has a mean of 98 and a standard deviation of 6. what is $p(x \leq 104)$?

a. 0.84

b. 0.975

c. 0.16

d. 0.025

Explanation:

Step1: Calculate the z-score

The formula for the z - score is $z=\frac{x - \mu}{\sigma}$, where $\mu$ is the mean, $\sigma$ is the standard deviation, and $x$ is the value from the dataset. Given $\mu = 98$, $\sigma=6$, and $x = 104$. Substitute these values into the formula: $z=\frac{104 - 98}{6}=\frac{6}{6}=1$.

Step2: Use the empirical rule for normal distribution

The empirical rule (68 - 95 - 99.7 rule) states that for a normal distribution:

• Approximately 68% of the data lies within $\mu\pm\sigma$ (z - scores between - 1 and 1).
• Approximately 95% of the data lies within $\mu\pm2\sigma$ (z - scores between - 2 and 2).
• Approximately 99.7% of the data lies within $\mu\pm3\sigma$ (z - scores between - 3 and 3).

We know that the total area under the normal curve is 1. The area to the left of $z = 1$ can be found using the fact that the area to the left of $\mu$ (z = 0) is 0.5, and the area between $z = 0$ and $z = 1$ is approximately 0.34 (since 68% of the data is between $z=- 1$ and $z = 1$, so half of that is 34% or 0.34). So $P(x\leq104)=P(z\leq1)=0.5 + 0.34=0.84$? Wait, no, actually, the empirical rule for z - score of 1: The cumulative probability for $z = 1$ (using standard normal tables or empirical rule) - the empirical rule more precisely, the area to the left of $z = 1$ is about 0.8413, which is approximately 0.84. But wait, another way: The area to the right of $z = 1$ is about 0.1587 (since the total area is 1, and the area between $z=-1$ and $z = 1$ is 0.68, so the area to the right of $z = 1$ is $\frac{1 - 0.68}{2}=0.16$), so the area to the left of $z = 1$ is $1 - 0.16 = 0.84$. Also, if we consider the 95% rule, for $z = 2$, the area to the left is 0.975, but here $z = 1$. Wait, maybe I made a mistake earlier. Wait, let's re - check. The z - score is 1. The standard normal table (or empirical rule) tells us that $P(Z\leq1)\approx0.8413\approx0.84$. But wait, the options have 0.975 which is for $z = 2$. Wait, no, wait, 104 is $\mu+\sigma$ (98 + 6=104), so $z = 1$. The area to the left of $z = 1$ is approximately 0.84. But wait, let's think again. Wait, the mean is 98, standard deviation 6. 104 is 98+6, so one standard deviation above the mean. The empirical rule says that 68% of data is between $\mu-\sigma$ and $\mu+\sigma$, so the area below $\mu+\sigma$ is $0.5+\frac{0.68}{2}=0.5 + 0.34 = 0.84$. So $P(x\leq104)$ is the area to the left of $x = 104$, which is the area to the left of $z = 1$, which is approximately 0.84. But wait, the option B is 0.975, which is for $z = 2$ (since $\mu + 2\sigma=98+12 = 110$). Wait, maybe I miscalculated the z - score? Wait, 104 - 98 is 6, divided by 6 is 1. So z - score is 1. So the probability should be approximately 0.84, which is option A? Wait, no, wait, maybe the question is using the 95% rule in a wrong way? Wait, no, let's check the standard normal distribution table. The cumulative probability for $z = 1$ is 0.8413, which is approximately 0.84. So the answer should be A. But wait, maybe I made a mistake. Wait, the mean is 98, standard deviation 6. 104 is 98 + 6, so z = 1. The area to the left of z = 1 is about 0.84. So the correct option is A.

Answer:

A. 0.84