Find the following probabilities based on the standard normal variable Z. (You may find it useful to reference the z table. Round your answers to 4 decimal places.)
P(-1.01<=Z<=-0.71)
P(0.01<=Z<=0.25)
P(-1.38<=Z<=0.06)
P(Z>3.2)
Using the z-table, we can find the probabilities for each of the given values.
P(-1.01<=Z<=-0.71):
Using the z-table, we find that the area to the left of -1.01 is 0.1562 and the area to the left of -0.71 is 0.2389. Therefore, the probability is P(-1.01<=Z<=-0.71) = 0.2389 - 0.1562 = 0.0827.
P(0.01<=Z<=0.25):
Using the z-table, we find that the area to the left of 0.01 is 0.5040 and the area to the left of 0.25 is 0.5987. Therefore, the probability is P(0.01<=Z<=0.25) = 0.5987 - 0.5040 = 0.0947.
P(-1.38<=Z<=0.06):
Using the z-table, we find that the area to the left of -1.38 is 0.0838 and the area to the left of 0.06 is 0.5239. Therefore, the probability is P(-1.38<=Z<=0.06) = 0.5239 - 0.0838 = 0.4401.
P(Z>3.2):
Using the z-table, we find that the area to the left of 3.2 is 0.9993. Therefore, the probability is P(Z>3.2) = 1 - 0.9993 = 0.0007.
To find the probabilities based on the standard normal variable Z, we can use the z-table.
1. P(-1.01 <= Z <= -0.71):
Subtracting the cumulative probability of -0.71 from the cumulative probability of -1.01, we have:
P(-1.01 <= Z <= -0.71) = P(Z <= -0.71) - P(Z <= -1.01)
2. P(0.01 <= Z <= 0.25):
Subtracting the cumulative probability of 0.01 from the cumulative probability of 0.25, we have:
P(0.01 <= Z <= 0.25) = P(Z <= 0.25) - P(Z <= 0.01)
3. P(-1.38 <= Z <= 0.06):
Subtracting the cumulative probability of 0.06 from the cumulative probability of -1.38, we have:
P(-1.38 <= Z <= 0.06) = P(Z <= 0.06) - P(Z <= -1.38)
4. P(Z > 3.2):
Since the z-table generally provides cumulative probabilities up to a certain value of Z, we need to find the area to the left of 3.2 and subtract it from 1:
P(Z > 3.2) = 1 - P(Z <= 3.2)
Using the z-table, we can look up the cumulative probabilities for the given values of Z and calculate the required probabilities.
To find these probabilities based on the standard normal variable Z, we can use the Z-table. The Z-table provides the cumulative probability up to a given Z-score.
1. P(-1.01<=Z<=-0.71):
First, look up the Z-score -1.01 in the left-hand column of the Z-table. Then find the corresponding value for -0.71 in the top row of the table. The intersection of these values gives you the cumulative probability to the left of -0.71, which is 0.2375. Next, look up the Z-score -1.01 in the left-hand column again, but this time find the corresponding value for -1.01 in the top row of the table. The intersection of these values gives you the cumulative probability to the left of -1.01, which is 0.1562. Now, subtract the cumulative probability to the left of -1.01 (0.1562) from the cumulative probability to the left of -0.71 (0.2375) to get the probability between -1.01 and -0.71: P(-1.01<=Z<=-0.71) = 0.2375 - 0.1562 = 0.0813.
2. P(0.01<=Z<=0.25):
Similarly, look up the Z-score 0.01 in the left-hand column of the Z-table and find the corresponding value for 0.01 in the top row. The intersection of these values gives you the cumulative probability to the left of 0.01, which is 0.5040. Next, look up the Z-score 0.25 in the left-hand column again, but this time find the corresponding value for 0.25 in the top row of the table. The intersection of these values gives you the cumulative probability to the left of 0.25, which is 0.5987. Now, subtract the cumulative probability to the left of 0.01 (0.5040) from the cumulative probability to the left of 0.25 (0.5987) to get the probability between 0.01 and 0.25: P(0.01<=Z<=0.25) = 0.5987 - 0.5040 = 0.0947.
3. P(-1.38<=Z<=0.06):
Use the same process as above to find the cumulative probabilities for Z=-1.38 and Z=0.06. Then subtract the cumulative probability to the left of -1.38 from the cumulative probability to the left of 0.06 to get the probability between -1.38 and 0.06.
4. P(Z>3.2):
To find the probability of Z being greater than 3.2, we need to find the cumulative probability to the left of 3.2 in the Z-table and subtract it from 1. Since the Z-table typically does not include values of Z greater than 3.09, we can assume that the probability of Z being greater than 3.2 is very close to 0. Therefore, P(Z>3.2) is approximately 0.
Remember to round your answers to 4 decimal places.