Anormal distribution has a mean of µ = 100 with σ = 20. if one score is randomly selected from this distribution, what is the probability that the score will have a value between x = 90 and x = 120?

0.5328 For this problem, you don't have a symmetric difference from the mean, but you can look at both sides. So first, figure out how many standard deviations you're below and above the mean. Standard deviations below the mean (90-100)/20 = -10/20 = -0.5 Standard deviations above the mean (120-100)/20 = 20/20 = 1.0 So your target ranges from half a deviation below to a deviation above the mean. Now using a standard normal table, look up the percentages associated with those deviations. Looking up -0.5 gives you a value of 0.19146 which means that 19.146% of the population lies between the mean and this deviation. Looking up 1.0 gives you a value of 0.34134 which means that 34.134% of the population lies between the mean and this deviation. Adding these values will give you the percent of the population that lies between the two limits. 0.19146 + 0.34134 = 0.5328 Therefore the probability of a random score in the population will have a value between 90 and 120 inclusive is 0.5328.

What is the inequality?

The answer to your question is d 68%