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Chapter 7: Problem 26

Find the \(Z\) -scores that separate the middle \(99 \%\) of the distribution from the area in the tails of the standard normal distribution. Find the \(Z\) -scores that separate the middle \(94 \%\) of the distribution from the area in the tails of the standard normal distribution.

Short Answer

Expert verified
The z-scores for the middle 99% are -2.576 and 2.576. The z-scores for the middle 94% are -1.881 and 1.881.

Step by step solution

01

Understand the problem

The goal is to find the z-scores that separate the middle 99% and 94% of the distribution from the tails of the standard normal distribution.
02

Calculate the tail areas

For the middle 99% of the distribution, the tails together will contain 1% of the area. Therefore, each tail will have 0.5% (or 0.005) of the area. For the middle 94% of the distribution, the tails together will contain 6% of the area, and each tail will have 3% (or 0.03) of the area.
03

Find the corresponding z-scores using z-tables

Look up the z-scores that correspond to the cumulative probabilities of 0.005 (for the lower tail of the middle 99%) and 0.995 (for the upper tail). Similarly, look up the z-scores for the cumulative probabilities of 0.03 (for the lower tail of the middle 94%) and 0.97 (for the upper tail).
04

Reference z-table values

From the z-table, the z-scores for the cumulative probabilities are: For 0.005: -2.576 For 0.995: 2.576. For 0.03: -1.881 For 0.97: 1.881
05

Conclusion

The z-scores stretching the middle 99% are -2.576 and 2.576. The z-scores stretching the middle 94% are -1.881 and 1.881.

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Key Concepts

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

Z-scores
A Z-score, also known as a standard score, measures how many standard deviations an element is from the mean. In a standard normal distribution, the mean is 0, and the standard deviation is 1. Z-scores are used to determine the position of a value within the distribution. If a Z-score is positive, the value is above the mean, and if it's negative, the value is below the mean. For example, a Z-score of 1.5 indicates the value is 1.5 standard deviations above the mean.
Cumulative Probability
Cumulative probability refers to the probability that a random variable is less than or equal to a certain value. In the context of the standard normal distribution, it is the area under the curve to the left of a particular Z-score. To find cumulative probabilities, we often use Z-tables. These tables provide the cumulative probability associated with each Z-score. For instance, a cumulative probability of 0.995 means 99.5% of the values lie below the corresponding Z-score.
Z-tables
Z-tables, or standard normal tables, are used to look up the cumulative probability associated with a Z-score in a standard normal distribution. They help us determine the probability of a score occurring within a given range. Using a Z-table, you can find the Z-score that corresponds to a specific cumulative probability. For example, to find the Z-score for a cumulative probability of 0.995, you look up this probability in the Z-table and find that it corresponds to a Z-score of approximately 2.576.
Tail Area
Tail areas are the regions at the far ends of the distribution curve. In the standard normal distribution, these areas represent the extreme values that are less likely to occur. To find the Z-scores that separate the tails from the middle of the distribution, we must calculate the tail area and then use the Z-table to find the corresponding Z-scores. For example, with a tail area of 0.005, each tail encompasses 0.5% of the total area, and the corresponding Z-scores are -2.576 and 2.576 for the lower and upper tails, respectively.

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Most popular questions from this chapter

Times The mean of the incubation time of fertilized chicken eggs kept at \(100.5^{\circ} \mathrm{F}\) in a still-air incubator is 21 days. Suppose the incubation times are approximately normally distributed with a standard deviation of 1 day. (Source: University of Illinois Extension.) (a) Determine the 17 th percentile for incubation times of fertilized chicken eggs. (b) Determine the incubation times that make up the middle \(95 \%\) of fertilized chicken eggs?

As reported by the U.S. National Center for Health Statistics, the mean height of females 20 to 29 years old is \(\mu=64.1\) inches. If height is approximately normally distributed with \(\sigma=2.8\) inches, answer the following questions: (a) What is the percentile rank of a 20 - to 29 -year-old female who is 60 inches tall? (b) What is the percentile rank of a 20 - to 29 -year-old female who is 70 inches tall? (c) What proportion of 20 - to 29 -year-old females are between 60 and 70 inches tall? (d) Would it be unusual for a 20 - to 29 -year-old female to be taller than 70 inches?

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?

Find the indicated \(Z\) -score. Be sure to draw a standard normal curve that depicts the solution. Find the \(Z\) -score such that the area under the standard normal curve to the right is 0.35

Find the \(Z\) -scores that separate the middle \(70 \%\) of the distribution from the area in the tails of the standard normal distribution. Find the \(Z\) -scores that separate the middle \(99 \%\) of the distribution from the area in the tails of the standard normal distribution.
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