6.2.3 Given a standardized normal distribution (with a mean of 0 and a standard

Question

6.2.3 Given a standardized normal distribution (with a mean of 0 and a standard deviation of 1), complete parts (a) through (d). EEB Click here to view page 1 of the cumulative standardized normal distribution table. E Click here to view page 2 of the cumulative standardized normal distribution table. a. What is the probability that Z is less than 1.01? The probability that Z is less than 1.01 is 0.8438 Round to four decimal places as needed.) b. What is the probability that Z is greater than -0.21? The probability that Z is greater than 0.21 is 0.5832 Round to four decimal places as needed.) c. What is the probability that Z is less than 0.21 or greater than the mean? The probability that Z is less than-0.21 or greater than the mean is Round to four decimal places as needed.)

Explanation / Answer

Part c:

P(Z less than -0.21 or greater than the mean) = This is basically equal to the union of these two probabilities.

We know that P(A union B) = P(A) + P(B) -P(A intersection B)

Therefore,

P(Z less than -0.21 or greater than the mean) = P(less than -0.21) +  P(greater than the mean) - P(greater than mean and less than -0.21)

According to the question, mean = 0.

So we can write the above equation as:

P(Z< -0.21 or Z>0) = P(Z<-0.21) +  P(Z>0) - P(Z>0 & Z< -0.21)

But there is no such Z which is greater than zero and less than -0.21.

So we can say that their intersection is zero.
Hence, the above equation becomes:

P(Z< -0.21 or Z>0) = P(Z<-0.21) +  P(Z>0) ............ (1)

Now, P(Z<-0.21) = 1 - P(Z>-0.21)

And probability of Z greater than -0.21,i.e., P(Z>-0.21), this you have already calculated in the part (b) of this question, whose value is 0.5832

Therefore, we get P(Z<-0.21) = 1 - 0.5832 = 0.4168

And we know that P(Z>0) = 0.5

Substituting these values in the equation (1), we get:
P(Z< -0.21 or Z>0) = P(Z<-0.21) +  P(Z>0) = 0.4168 + 0.5 = 0.9168

Answer = 0.9168

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