# For a variable that is normally distributed with a mean of 50 and a standard deviation of 5, approximately _? of the scores are between _?. A)95%, 45 and 55 B)95%, 40 and 60 C)68%, 40 and 60 D)95%, 35 and 65 E)99%, 40 and 60

For a variable that is normally distributed with a mean of 50 and a standard deviation of 5, approximately _? of the scores are between _?.
A)95%, 45 and 55
B)95%, 40 and 60
C)68%, 40 and 60
D)95%, 35 and 65
E)99%, 40 and 60
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Step 1
Empirical rule:
For symmetric or normal distributions,
68% of values fall within one standard deviation from the mean. That is, (m – s, m + s).
95% of values fall within two standard deviations from the mean. That is, (m – 2*s, m + 2*s).
7% of values fall within three standard deviations from the mean. That is, (m – 3*s, m + 3*s).
Step 2
Calculations:
The random variable x follows normal distribution with mean 50 and standard deviation 5.
That is, m = 50 and s = 5.
The calculation based on empirical rule is given below:
One $\sigma$ limits:
$\left(\mu ±\sigma \right)=\left(5--5,50+5\right)=\left(45,55\right)$
68% of scores are between 45 and 55.
Two sigma limits:
$\left(\mu ±2\sigma \right)=\left(50-2×5,50+2×5\right)=\left(40,60\right)$
95% of scores are between 40 and 60.
Three $\sigma$ limits:
$\left(\mu ±3\sigma \right)=\left(50-3×5,50+3×5\right)=\left(35,65\right)$
99% of scores are between 35 and 65.
Step 3
Obtain the correct option:
From the given 5 options, option (B) states that 95% of the scores lie between 40 and 60.
Hence, option (B) is correct.
Step 4
Option (B) is correct.