Question
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)
P(-1.01<=Z<=-0.71)
P(0.01<=Z<=0.25)
P(-1.38<=Z<=0.06)
P(Z>3.2)
Answers
GPT 3.5
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.
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.