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.

I get 1.89 also.

Not sure if there is a typo or I made a mistake.

To solve this problem, we can use the standard normal distribution table, also known as the z-table.

First, let's find the z-score for the value 12.10 using the formula:
z = (x - μ) / σ
where z is the z-score, x is the value, μ is the mean, and σ is the standard deviation.

z = (12.10 - 11) / σ

The probability that X is less than 12.10 is given as 0.72. In terms of z-scores, this probability can be written as P(Z < z) = 0.72.

Now, we can use the z-table to find the z-score that corresponds to a cumulative probability of 0.72. From the z-table, we find that the z-score is approximately 0.618.

0.618 = (12.10 - 11) / σ

Now, we can solve for σ:

0.618σ = 1.10

σ = 1.10 / 0.618 = 1.779

So, the standard deviation of X is approximately 1.779, which is not provided as one of the answer choices.

It seems there may be an error in either the question or the provided answer choices. It is always important to double-check the information given and the options available.