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Use the Standard Normal Distribution table to find theindicated area under the s

ID: 2919240 • Letter: U

Question

Use the Standard Normal Distribution table to find theindicated area under the standard normal curve.

Q1: Between z = 0 and z = 2.24
A1: 0 = 0.5000
       2.24 = 0.9875
       0.9875 – 0.5000 =0.4875

Q2:To the left of z = 1.09
A2: 0.8621

Q3: Between z = -1.15 and z = -0.56
A3: -1.15 = 0.1251
       -0.56 = 0.2877
       0.1251 - 0.2877 = -0.1626

Q4: To the right of z = -1.93
A4: 1 – 0.0268 = 0.9732
Section 5.2: Normal Distributions: Find Probabilities

Q5: The diameters of a wooden dowel produced by a new machine arenormally distributed with a mean of 0.55 inches and a standarddeviation of 0.01 inches. What percent of the dowels will have adiameter greater than 0.57?

A5: z = x - µ / ó
          = 0.57– 0.55 / 0.01 = 2
         =P (x >0.57)
         = P(z >2)
         = 1 – P(z < 2)
         = 1 –0.9772
         = 0.0228

Q6: A loan officer rates applicants for credit. Ratings arenormally distributed. The mean is 240 and the standard deviation is50. Find the probability that an applicant will have a ratinggreater than 260.

A6: z = x - µ / ó
          = 260– 240 / 50
          = 0.4
         =P (x >260)
         = P(z >0.4)
        = 1 – P (z< 0.4)
         = 1 –0.6554
         = 0.9772

Use the Standard Normal Distribution table to find theindicated area under the standard normal curve.

Q1: Between z = 0 and z = 2.24
A1: 0 = 0.5000
       2.24 = 0.9875
       0.9875 – 0.5000 =0.4875

Q2:To the left of z = 1.09
A2: 0.8621

Q3: Between z = -1.15 and z = -0.56
A3: -1.15 = 0.1251
       -0.56 = 0.2877
       0.1251 - 0.2877 = -0.1626

Q4: To the right of z = -1.93
A4: 1 – 0.0268 = 0.9732
Section 5.2: Normal Distributions: Find Probabilities

Q5: The diameters of a wooden dowel produced by a new machine arenormally distributed with a mean of 0.55 inches and a standarddeviation of 0.01 inches. What percent of the dowels will have adiameter greater than 0.57?

A5: z = x - µ / ó
          = 0.57– 0.55 / 0.01 = 2
         =P (x >0.57)
         = P(z >2)
         = 1 – P(z < 2)
         = 1 –0.9772
         = 0.0228

Q6: A loan officer rates applicants for credit. Ratings arenormally distributed. The mean is 240 and the standard deviation is50. Find the probability that an applicant will have a ratinggreater than 260.

A6: z = x - µ / ó
          = 260– 240 / 50
          = 0.4
         =P (x >260)
         = P(z >0.4)
        = 1 – P (z< 0.4)
         = 1 –0.6554
         = 0.9772

Explanation / Answer

Q3: Between z = -1.15 and z = -0.56 A3: -1.15 = 0.1251        -0.56 = 0.2877        0.1251 -0.2877 = -0.1626(wrong)        0.2877-0.1251=0.1626 Others are correct.