Question

suppose a normal distribution has a mean of 222 and a standard deviation of 16. what is the probability that a data value is between 174 and 238? round your answer to the nearest tenth of a percent.

choice

a. 84.0%
b. 91.0%
c. 66.9%
d. 53.3%

Explanation:

Step1: Calculate z - scores

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 data value.

For $x = 174$:
$z_1=\frac{174 - 222}{16}=\frac{- 48}{16}=- 3$

For $x = 238$:
$z_2=\frac{238 - 222}{16}=\frac{16}{16}=1$

Step2: Find probabilities from z - table

We know that for a standard normal distribution:

• $P(Z < - 3)\approx0.0013$ (from z - table, the area to the left of $z=-3$)
• $P(Z < 1)\approx0.8413$ (from z - table, the area to the left of $z = 1$)

The probability that a data value is between 174 and 238 is $P(-3

Substitute the values: $P(-3

• Approximately 68% of data is within 1 standard deviation of the mean ($\mu\pm\sigma$)
• Approximately 95% within 2 standard deviations
• Approximately 99.7% within 3 standard deviations.

Wait, but let's check the z - scores again. Wait, maybe I miscalculated the upper z - score. Wait, 238-222 = 16, 16/16 = 1. Correct. 174 is 222 - 48, 48/16 = 3. Correct. So $P(-316=238, correct. 222-316 = 174, correct. So the area between $z=-3$ and $z = 1$ is 0.8413 - 0.0013=0.84, so 84.0%, which is option A.

Wait, but maybe the question is using a different approach? Wait, let's re - check. The mean is 222, standard deviation 16.

Data value 174: $z=(174 - 222)/16=-3$

Data value 238: $z=(238 - 222)/16 = 1$

We need $P(174

From standard normal table:

$P(Z < 1)=0.8413$

$P(Z < - 3)=0.0013$

So $P(-3

Answer:

A. 84.0%