Question

According to Experian, the average credit score of and American adult is 705 and the standard deviation of all credit scores of American adults is 64. Determine the following probabilities. Round solutions to four decimal places, if necessary. Determine the probability that a randomly selected American adult has a credit score less than or equal to 834. P(x≤834)= Determine the probability that a randomly selected American adult has a credit score between 679 and 917. P(679

Answer 1

1. P(x ≤ 834) = 0.9706 (approximately)

2. P(679 < x < 917) = 0.8370 (approximately)

Step-by-step explanation:

1. To determine the probability that a randomly selected American adult has a credit score less than or equal to 834, we'll use z-score formula:
`\(Z = (X - \mu)/(\sigma)\),` where \(X\) is the value (834), \(\mu\) is the mean (705), and \(\sigma\) is the standard deviation (64). Calculate the z-score: \(Z = \frac{834 - 705}{64} = 2.0156\). Using a standard normal distribution table or calculator, find the probability corresponding to \(Z = 2.0156\), which is approximately 0.9706.

2. For the probability that a randomly selected American adult has a credit score between 679 and 917, first, determine the z-scores for these values
`: \(Z_(679) = (679 - 705)/(64) = -0.4063\) and \(Z_(917) = (917 - 705)/(64) = 3.3125\).`Then find the area between these z-scores in the standard normal distribution table or using a calculator. Subtract the cumulative probability at \(Z = -0.4063\) from the cumulative probability at \(Z = 3.3125\) to get the probability between these two values, which is approximately 0.8370.

These probabilities are calculated using the properties of the standard normal distribution and z-scores to determine the likelihood of a randomly chosen American adult having a credit score within specific ranges.