A normal random variable X has mean 5.6 and standard deviation 1.2. For which c> 0 does the equality P(-c≤x≤c) 0.9 hold? Round to two decimal places if needed.

To solve this problem, we need to standardize the normal random variable X. We do this by subtracting the mean and dividing by the standard deviation. This gives us Z = (X - 5.6) / 1.2.

The probability P(-c ≤ X ≤ c) = 0.9 is equivalent to P(-c ≤ Z ≤ c) = 0.9 after standardization.

Since the total probability under the standard normal curve is 1, and we know that P(-c ≤ Z ≤ c) = 0.9, this means that the total probability in the tails of the distribution is 1 - 0.9 = 0.1.

Because the standard normal distribution is symmetric, the probability in each tail of the distribution is 0.1 / 2 = 0.05.

We can now use a standard normal distribution table or a Z-score calculator to find the value of c that corresponds to a tail probability of 0.05. This value is approximately 1.645.

However, this is the standardized value. To get the value of c for the original variable X, we need to transform it back using the formula c = Z * standard deviation + mean.

Substituting the values we have, c = 1.645 * 1.2 + 5.6 = 7.574.

So, for c > 0, the equality P(-c ≤ X ≤ c) = 0.9 holds when c is approximately 7.57 (rounded to two decimal places).