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the interest rate on a five-year certificate of deposit (cd) for banks in south florida in 2018 was normally distributed with a mean of 2.28 percent and a standard deviation of 0.37 percent. if a bank has an interest rate at the 45th percentile, what is the interest rate at this bank?



Answer :

The interest rate at the bank = 2.324 %

Let X be the random variable representing the interest rate on a five-year certificate of deposit (cd) for banks in South Florida in 2018.

Then X follows the Normal Distribution with,

Mean, μ = 2.28 %  and,

Standard Deviation, σ = 0.37 %

We need to find the X at the 45th percentile.

45th percentile implies z-value = 0.45

The z-score at 45th percentile or 0.45 probability = -0.12 [From the z-tables]

We have, z = (x-μ)/σ

⇒ zσ = x-μ

⇒ x = zσ+μ

⇒ x = (0.12 x 0.37) + 2.28

      = 2.324

Hence the interest rate at the bank = 2.324 %

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