A bank's loan officer rates applicants for credit. The ratings are normally distributed with a mean of 200 and a standard deviation of 50. If an applicant is randomly selected what is the probability of a rating that is between 225 and 276

P(225 ≤ x ≤ 276) = 0.25

Step-by-step explanation:

Since the credit ratings for applicants are normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = credit ratings of applicants

µ = mean

σ = standard deviation

From the information given,

µ = 200

σ = 50

The probability of a rating that is between 225 and 276 is expressed as

P(225 ≤ x ≤ 276)

For x = 225,

z = (225 - 200)/50 = 0.5

Looking at the normal distribution table, the probability corresponding to the z score is 0.69

For x = 276,

z = (276 - 200)/50 = 1.52

Looking at the normal distribution table, the probability corresponding to the z score is 0.94

Therefore,

P(225 ≤ x ≤ 276) = 0.94 - 0.69 = 0.25

2

step-by-step explanation: