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, find the probability of a rating that is between 200 and 275.

Answer:

0.4332 = 43.32% probability of a rating that is between 200 and 275.

Step-by-step explanation:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

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

This means that [tex]\mu = 200, \sigma = 50[/tex].

Find the probability of a rating that is between 200 and 275.

This is the pvalue of Z when X = 275 subtracted by the pvalue of Z when X = 200. So

X = 275

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{275 - 200}{50}[/tex]

[tex]Z = 1.5[/tex]

[tex]Z = 1.5[/tex] has a pvalue of 0.9332

X = 200

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{200 - 200}{50}[/tex]

[tex]Z = 0[/tex]

[tex]Z = 0[/tex] has a pvalue of 0.5

0.9332 - 0.5 = 0.4332

0.4332 = 43.32% probability of a rating that is between 200 and 275.

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