Question

A random variable is normally distributed with a mean of μμ = 50 and a standard deviation of σσ = 5. What is the probability (NOT percentage) the random variable will assume a value between 45 and 55 (to 4 decimals)? Do not convert to percent. Hint: Consider that 5 is one standard deviation from the mean of 50.

Answer 1

Answer:

0.6827

Step-by-step explanation:

In a normal distribution with mean $\bf \mu$ and standard deviation $\bf \sigma$, 68.27% of the data fall between $\bf \mu-\sigma$ and $\bf \mu+\sigma$  

In this case $\bf \mu=50$, $\bf \sigma=5$, so 68.27% of the data fall between 45 and 55.

Hence, the probability (not percentage) the random variable will assume a value between 45 and 55 (to 4 decimals) is

0.6827

Answer 2

The probability that the random variable will assume a value between 45 and 55 is given by: 0.6827

Given that:

• A random variable is normally distributed.
• Mean μ = 50
• Standard deviation σ = 5

What is empirical rule?

According to the empirical rule, also known as 68-95-99.7 rule, the percentage of values that lie within an interval with 68%, 95% and 99.7% of the values lies within one, two or three standard deviations of the mean of the distribution.

The more precise statement for 68 percent is:

$P(\mu -\sigma < X < \mu + \sigma ) \approx 68.27\%$

Since the interval of interest given in question is (45,55), it can be rewritten as  (50-5, 50 + 5)

Thus we have:

$P(50-5 < X < 50+5) \approx 68.27\%\\P(45 < X < 55) \approx 68.27\%\\P(45 < X < 55) \approx 0.6827\: \rm probability$

Thus, 0.6827 is the probability that the random variable will assume a value between 45 and 55.

Learn more about such probability here:

https://brainly.com/question/1804405