Final answer:
The probability that all 5 values from a standard normal distribution are greater than -2.2 is about 0.9313, calculated by raising the CDF value of a single value being greater than -2.2 (approximately 0.9861) to the power of 5.
Step-by-step explanation:
The question asks for the probability that all 5 values, drawn independently from the standard normal distribution, are greater than -2.2. This is a problem involving the cumulative distribution function (CDF) of the standard normal distribution.
First, we find the CDF value for a single value being greater than -2.2. The z score of -2.2 corresponds to a large cumulative probability, since most of the standard normal distribution lies to the right of this point. Using a standard normal distribution table or a calculator, we find that the cumulative probability for a z score greater than -2.2 is approximately 0.9861.
Since the values are drawn independently, the joint probability of all 5 values being greater than -2.2 is the product of their individual probabilities: P(X1 > -2.2, X2 > -2.2, ..., X5 > -2.2) = P(X > -2.2)^5, where X represents a standard normal random variable.
Therefore, the probability that all 5 values are greater than -2.2 is about 0.9861^5 = 0.9313 to three decimal places.