Use the Standard Normal Distribution table to find the indicated 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 are normally distributed with a mean of 0.55 inches and a standard deviation of 0.01 inches. What percent of the dowels will have a diameter 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 are normally distributed. The mean is 240 and the standard deviation is 50. Find the probability that an applicant will have a rating greater 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
1. Right, but there should be a column in the table listing area from mean to Z score, so you could have eliminated some of the steps you indicate.
4, 5, 6. Likewise, there should be a column indicating the area in the larger portion and smaller portion to find these answers more directly.
6. You have a subtraction error. 1 - 0.6554 ≠ 0.9772.
Otherwise, you are correct.
Q7: Why did the scarecrow win an award?
A7: Because he was outstanding in his field!
Q1: Between z = 0 and z = 2.24
A1: The area between z = 0 and z = 2.24 is 0.4875.
Q2: To the left of z = 1.09
A2: The area to the left of z = 1.09 is 0.8621.
Q3: Between z = -1.15 and z = -0.56
A3: The area between z = -1.15 and z = -0.56 is -0.1626.
Q4: To the right of z = -1.93
A4: The area to the right of z = -1.93 is 0.9732.
Q5: The diameters of a wooden dowel produced by a new machine are normally distributed with a mean of 0.55 inches and a standard deviation of 0.01 inches. What percent of the dowels will have a diameter greater than 0.57?
A5: The probability of a wooden dowel having a diameter greater than 0.57 is 0.0228.
Q6: A loan officer rates applicants for credit. Ratings are normally distributed. The mean is 240 and the standard deviation is 50. Find the probability that an applicant will have a rating greater than 260.
A6: The probability of an applicant having a rating greater than 260 is 0.9772.
In order to find the indicated area under the standard normal curve, we need to use the Standard Normal Distribution Table (also known as the z-table). This table provides the cumulative probability (area) to the left of a given z-score.
To use the table, follow these steps:
1. Look for the desired z-score in the leftmost column of the table.
2. Look for the desired decimal value in the top row of the table. The decimal values represent the second digit of the z-score rounded to two decimal places.
3. The intersection of the row and column will give you the cumulative probability (area) to the left of the z-score.
For example, let's use the table to find the area between z = 0 and z = 2.24 (Q1). First, we look for 0 in the leftmost column and find the corresponding value of 0.5000 in the top row. This represents the cumulative probability to the left of z = 0.
Next, we look for 2.24 in the leftmost column and find the corresponding value of 0.9875 in the top row. This represents the cumulative probability to the left of z = 2.24.
To find the area between z = 0 and z = 2.24, we subtract the cumulative probabilities: 0.9875 - 0.5000 = 0.4875.
For the other questions, follow the same steps using the appropriate z-scores and values from the table.
For Q2, we want to find the area to the left of z = 1.09. We find the corresponding cumulative probability in the table, which is 0.8621.
For Q3, we want to find the area between z = -1.15 and z = -0.56. The corresponding cumulative probabilities are 0.1251 and 0.2877. By subtracting these values, we get 0.1626.
For Q4, we want to find the area to the right of z = -1.93. The table provides the cumulative probability to the left of z, so to find the area to the right, we subtract the cumulative probability from 1. In this case, it is 1 - 0.0268 = 0.9732.
For Q5 and Q6, we need to calculate the z-scores first using the formula z = (x - µ) / σ, where x is the given value, µ is the mean, and σ is the standard deviation. Then, we use the table to find the cumulative probability and subtract it from 1 to find the area to the right.
In Q5, the z-score is 2. We find the cumulative probability to the left of z = 2 in the table, which is 0.9772. To find the area to the right, we subtract it from 1: 1 - 0.9772 = 0.0228.
In Q6, the z-score is 0.4. The cumulative probability to the left of z = 0.4 is 0.6554. To find the area to the right, we subtract it from 1: 1 - 0.6554 = 0.3446.