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Chapter 6: Problem 26

Obtain the following probabilities for the standard normal distribution. a. \(P(z>-1.86)\) b. \(P(-.68 \leq z \leq 1.94)\) c. \(P(0 \leq z \leq 3.85)\) d. \(P(-4.34 \leq z \leq 0)\) e. \(P(z>4.82)\) f. \(P(z<-6.12)\)

Short Answer

Expert verified
The following are the probabilities for the given z-scores: \(P(z>-1.86)=0.9686\), \(P(-.68 \leq z \leq 1.94)=0.7261\), \(P(0 \leq z \leq 3.85)=0.4999\), \(P(-4.34 \leq z \leq 0)=0.5\), \(P(z>4.82)\approx 0\), and \(P(z<-6.12)\approx 0\).

Step by step solution

01

Understanding Z-score Probabilities

The first thing to understand is how z-scores translate to probabilities. For a given z-score, the associated probability equals the area under the standard normal curve to the left of that z-score. So to find the probability that the z-score is less than a certain value, simply look up the corresponding area in the standard normal table. To find the probability that the z-score lies within a certain range, subtract the smaller z-score's area from the larger z-score's area.
02

Calculating Probability for \(P(z > -1.86)\)

To calculate \(P(z > -1.86)\), we first look up the z-score in the standard normal table. The area to the left of -1.86 is approximately 0.0314. However, we want the probability to the right of -1.86, so we subtract this area from 1. Therefore, \(P(z > -1.86) = 1 - 0.0314 = 0.9686\).
03

Calculating Probability for \(P(-.68 \leq z \leq 1.94)\)

We first find the area to the left of -0.68, which is approximately 0.2483, then the area to the left of 1.94 which is approximately 0.9744. Subtracting these gives the probability that z lies between these two values. Therefore, \(P(-.68 \leq z \leq 1.94) = 0.9744 - 0.2483 = 0.7261\).
04

Calculating Probability for \(P(0 \leq z \leq 3.85)\)

Find the area to the left of 0 which is 0.5, and the area to the left of 3.85 which is approximately 0.9999. Subtracting these gives the probability that z lies between these two values. So, \(P(0 \leq z \leq 3.85) = 0.9999 - 0.5 = 0.4999\).
05

Calculating Probability for \(P(-4.34 \leq z \leq 0)\)

Find the area to the left of -4.34, which is so small it is essentially 0, and the area to the left of 0, which is 0.5. Subtracting these gives the probability that z lies between -4.34 and 0. Therefore, \(P(-4.34 \leq z \leq 0) = 0.5 - 0 = 0.5\).
06

Calculating Probability for \(P(z > 4.82)\) and \(P(z < -6.12)\)

For both \(P(z > 4.82)\) and \(P(z < -6.12)\), since these represent z-values that are very far from the mean, the areas to their left and right in the standard normal table are very small -- so small that they are essentially 0. Therefore, both \(P(z > 4.82)\) and \(P(z < -6.12)\) are approximately equal to 0.

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Most popular questions from this chapter

The amount of time taken by a bank teller to serve a randomly selected customer has a normal distribution with a mean of 2 minutes and a standard deviation of \(.5\) minute. a. What is the probability that both of two randomly selected customers will take less than I minute each to be served? b. What is the probability that at least one of four randomly selected customers will need more than \(2.25\) minutes to be served?

Briefly describe the standard normal distribution curve.

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