If X is a normal random variable with mean 11, and if the probability that X is less than 12.10 is .72 , then what is the standard deviation of X?
Select one:
a. 3.00
b. 6.25
c. 2.50
d. 1.25
Please note that the answer for the above is stated as c.2.50. However, my answer comes to 1.89 which is not given as an option and I can't figure out where I'm going wrong so please can anyone help me with this problem.
Thanks.
You are correct. I also get 1.89 for the standard deviation.
If the standard deviation were 2.50, only 67% of X numbers would be below 12.10
Thank you!
To find the standard deviation of the normal random variable X, we need to use the concept of the standard normal distribution.
The standard normal distribution has a mean of 0 and a standard deviation of 1. We can standardize any normal random variable by subtracting its mean and dividing by its standard deviation.
In this case, we are given that X has a mean of 11. To standardize X, we subtract the mean from X and divide by the standard deviation. Let's call the standard deviation of X as σ.
Standardizing the variable X, we have:
Z = (X - 11) / σ
We are also given that the probability that X is less than 12.10 is 0.72. We can convert this to a standard normal distribution using the Z-score obtained by the standardization.
Using the standard normal distribution table or software, we can look up the Z-score that corresponds to a cumulative probability of 0.72. Let's call this Z-score Zα.
Zα = 0.72
Now we can solve for the standard deviation σ by plugging in the values:
Zα = (12.10 - 11) / σ
0.72 = 1.10 / σ
σ = 1.10 / 0.72
σ ≈ 1.53
However, option c (2.50) was clearly given as the correct answer. It seems there might be an error in either the information given or the answer choices themselves. It's always a good idea to double-check the question and answer choices to ensure accuracy.