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Find the indicated probability of the standard normal random variable \(Z\). $$P(Z<-2.56 \text { or } Z>1.39)$$

Short Answer

Expert verified
0.0875

Step by step solution

01

- Identify the given probabilities

We need to calculate the probability of the standard normal variable Z being less than -2.56 or greater than 1.39. Mathematically, this is represented as: \[ P(Z<-2.56 \text { or } Z>1.39) \]
02

- Use Z-table for left tail probability

First, find the probability that Z is less than -2.56. According to the Z-table, the probability that Z is less than -2.56 is approximately 0.0052. Therefore, \[ P(Z<-2.56) = 0.0052 \]
03

- Use Z-table for right tail probability

Next, find the probability that Z is greater than 1.39. The Z-table provides the cumulative probability for Z<1.39, which is approximately 0.9177. Therefore, the probability to the right is: \[ P(Z>1.39) = 1 - P(Z<1.39) = 1 - 0.9177 = 0.0823 \]
04

- Combine both probabilities

Now, sum the probabilities from Step 2 and Step 3 to get the total probability: \[ P(Z<-2.56 \text { or } Z>1.39) = P(Z<-2.56) + P(Z>1.39) = 0.0052 + 0.0823 = 0.0875 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Standard Normal Distribution
The standard normal distribution is a special case of the normal distribution. It has a mean of 0 and a standard deviation of 1. This distribution is symmetric about the mean, which means the left side mirrors the right side.
When working with the standard normal distribution, we use the variable Z to represent a data point. The value of Z tells us how many standard deviations away from the mean our data point lies.
If Z is positive, the data point is above the mean. If Z is negative, it is below the mean. Most values in a standard normal distribution fall within three standard deviations of the mean. Understanding this helps in calculating probabilities for given Z values.
Cumulative Probability
Cumulative probability is the probability that a random variable is less than or equal to a certain value. For the standard normal distribution, cumulative probability refers to the probability that Z is less than or equal to a certain Z value.
To find this, we use a Z-table which provides cumulative probabilities for various Z values. The table gives us the area under the curve to the left of a Z value. For example, if the Z value is 1.39, the cumulative probability is approximately 0.9177. This means there is a 91.77% chance that a randomly selected data point from the standard normal distribution will be less than 1.39.
In problems combining multiple Z values, such as finding the probability that Z is either less than -2.56 or greater than 1.39, we use cumulative probabilities for each part and then sum them up.
Z-Table Usage
The Z-table is a crucial tool for finding cumulative probabilities and solving problems related to the standard normal distribution. This table lists Z values and their corresponding cumulative probability (the area under the curve to the left of the Z value).
To use the Z-table, locate the Z value in the table. For Z values not listed, approximate by using the closest available value. For example, to find the probability of Z being less than -2.56, locate -2.56 in the Z-table, which shows a cumulative probability of 0.0052.
For Z values greater than the mean, like Z = 1.39, find the cumulative probability from the Z-table (0.9177). To find the probability Z is greater than 1.39, subtract this cumulative probability from 1 (since the total area under the curve is 1):
1 - 0.9177 = 0.0823
This means there’s an 8.23% chance that Z is greater than 1.39.
The Z-table enables quick and accurate calculation of these kinds of probabilities, making it invaluable for statistics and probability exercises.

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Most popular questions from this chapter

Find the indicated probability of the standard normal random variable \(Z\). $$P(Z>0.92)$$

The number of chocolate chips in an 18 -ounce bag of Chips Ahoy! chocolate chip cookies is approximately normally distributed with a mean of 1262 chips and standard deviation 118 chips according to a study by cadets of the U.S. Air Force Academy. (Source: Brad Warner and Jim Rutledge, Chance, Vol. 12, No. \(1,1999,\) pp. \(10-14 .\) ) (a) What is the probability that a randomly selected 18 ounce bag of Chips Ahoy! cookies contains between 1000 and 1400 chocolate chips? (b) What is the probability that a randomly selected 18 ounce bag of Chips Ahoy! cookies contains fewer than 1000 chocolate chips? (c) What proportion of 18 -ounce bags of Chip Ahoy! cookies contains more than 1200 chocolate chips? (d) What proportion of 18 -ounce bags of Chip Ahoy! cookies contains fewer than 1125 chocolate chips? (e) What is the percentile rank of an 18 -ounce bag of Chip Ahoy! cookies that contains 1475 chocolate chips? (f) What is the percentile rank of an 18-ounce bag of Chip Ahoy! cookies that contains 1050 chocolate chips?

The length of human pregnancies are approximately normally distributed with mean \(\mu=266\) days and standard deviation \(\sigma=16\) days. (a) What percent of pregnancies lasts more than 270 days? (b) What percent of pregnancies lasts less than 250 days? (c) What percent of pregnancies lasts between 240 and 280 days? (d) What is the probability that a randomly selected pregnancy lasts more than 280 days? (e) What is the probability that a randomly selected pregnancy lasts no more than 245 days? (f) A "very preterm" baby is one whose gestation period is less than 224 days. What proportion of births is "very preterm"?

Find the indicated probability of the standard normal random variable \(Z\). $$P(Z \leq-2.69)$$

The birth weights of full-term babies are normally distributed with mean \(\mu=3400\) grams and \(\sigma=505\) grams. (Source: Based on data obtained from the National Vital Statistics Report, Vol. 48, No.3) (a) Draw a normal curve with the parameters labeled. (b) Shade the region that represents the proportion of full-term babies who weigh more than 4410 grams. (c) Suppose the area under the normal curve to the right of \(X=4410\) is \(0.0228 .\) Provide two interpretations of this result.

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