Solved on Dec 13, 2023

Find the probability that a standard normal random variable zz is less than 0.24 or greater than or equal to -0.42.

STEP 1

Assumptions
1. We are dealing with a standard normal distribution, which is symmetric about the mean, z=0z=0.
2. We want to find the probability that zz is either less than 0.240.24 or greater than or equal to 0.42-0.42.
3. The probability of any event in a standard normal distribution can be found using the standard normal distribution table or a calculator with statistical functions.

STEP 2

The probability that zz is less than 0.240.24 can be found directly using the standard normal distribution table or statistical functions on a calculator.
P(z<0.24)P(z < 0.24)

STEP 3

To find P(z<0.24)P(z < 0.24), we look up the value of 0.240.24 in the standard normal distribution table or use a statistical calculator.

STEP 4

The probability that zz is greater than or equal to 0.42-0.42 can be found by subtracting the probability that zz is less than 0.42-0.42 from 11, since the total area under the standard normal curve is 11.
P(z0.42)=1P(z<0.42)P(z \geq -0.42) = 1 - P(z < -0.42)

STEP 5

To find P(z<0.42)P(z < -0.42), we look up the value of 0.42-0.42 in the standard normal distribution table or use a statistical calculator. Remember that due to the symmetry of the standard normal distribution, P(z<0.42)=P(z>0.42)P(z < -0.42) = P(z > 0.42).

STEP 6

Now we can calculate P(z0.42)P(z \geq -0.42) using the value found in the previous step.
P(z0.42)=1P(z<0.42)P(z \geq -0.42) = 1 - P(z < -0.42)

STEP 7

The probability that zz is either less than 0.240.24 or greater than or equal to 0.42-0.42 is the sum of the probabilities found in steps 3 and 6.
P(z<0.24 or z0.42)=P(z<0.24)+P(z0.42)P(z < 0.24 \text{ or } z \geq -0.42) = P(z < 0.24) + P(z \geq -0.42)

STEP 8

Add the two probabilities together to find the final answer.
P(z<0.24 or z0.42)=P(z<0.24)+(1P(z<0.42))P(z < 0.24 \text{ or } z \geq -0.42) = P(z < 0.24) + (1 - P(z < -0.42))

STEP 9

Assuming we have found the values from the standard normal distribution table or a statistical calculator:
Let's say P(z<0.24)=aP(z < 0.24) = a and P(z<0.42)=bP(z < -0.42) = b, then we can write:
P(z<0.24 or z0.42)=a+(1b)P(z < 0.24 \text{ or } z \geq -0.42) = a + (1 - b)

STEP 10

Now, plug in the values for aa and bb to get the numerical answer.
P(z<0.24 or z0.42)=a+(1b)P(z < 0.24 \text{ or } z \geq -0.42) = a + (1 - b)
For example, if a=0.595a = 0.595 and b=0.337b = 0.337, then:
P(z<0.24 or z0.42)=0.595+(10.337)P(z < 0.24 \text{ or } z \geq -0.42) = 0.595 + (1 - 0.337)

STEP 11

Calculate the final probability.
P(z<0.24 or z0.42)=0.595+0.663P(z < 0.24 \text{ or } z \geq -0.42) = 0.595 + 0.663
P(z<0.24 or z0.42)=1.258P(z < 0.24 \text{ or } z \geq -0.42) = 1.258
However, since the total probability cannot exceed 11, we must realize that there is an overlap in the events z<0.24z < 0.24 and z0.42z \geq -0.42. This overlap is exactly the probability P(0.42z<0.24)P(-0.42 \leq z < 0.24), which we have mistakenly counted twice.

STEP 12

To correct the double-counting, we subtract the overlap probability P(0.42z<0.24)P(-0.42 \leq z < 0.24) from the sum we calculated.
P(0.42z<0.24)=P(z<0.24)P(z<0.42)P(-0.42 \leq z < 0.24) = P(z < 0.24) - P(z < -0.42)
P(0.42z<0.24)=abP(-0.42 \leq z < 0.24) = a - b
P(0.42z<0.24)=0.5950.337P(-0.42 \leq z < 0.24) = 0.595 - 0.337
P(0.42z<0.24)=0.258P(-0.42 \leq z < 0.24) = 0.258

STEP 13

Now, subtract the overlap probability from the sum we calculated in step 11.
P(z<0.24 or z0.42)=1.2580.258P(z < 0.24 \text{ or } z \geq -0.42) = 1.258 - 0.258
P(z<0.24 or z0.42)=1.000P(z < 0.24 \text{ or } z \geq -0.42) = 1.000

STEP 14

The final probability is therefore 1.0001.000, which makes sense because the events z<0.24z < 0.24 and z0.42z \geq -0.42 cover the entire range of possibilities for a standard normal distribution.
P(z<0.24 or z0.42)=1.000P(z < 0.24 \text{ or } z \geq -0.42) = 1.000

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