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Chapter 6: Problem 25

Find the following probabilities for the standard normal distribution. a. \(P(z<-1.31)\) b. \(P(1.23 \leq z \leq 2.89)\) c. \(P(-2.24 \leq z \leq-1.19)\) d. \(P(z<2.02)\)

Short Answer

Expert verified
a) \(P(z<-1.31) = 0.0951\), b) \(P(1.23 鈮� z 鈮� 2.89) = 0.1074\), c) \(P(-2.24 鈮� z 鈮� -1.19)= 0.1045\), and d) \(P(z<2.02)= 0.9783\).

Step by step solution

01

Find P(z < -1.31)

Looking at the z-table, the area to the left of \(z = -1.31\) is found to be 0.0951. Therefore, \(P(z<-1.31) = 0.0951.\)
02

Find P(1.23 鈮� z 鈮� 2.89)

To find this probability, first find the area to the left of \(z = 2.89\) in the Z-table, which is approximately 0.9981. Now, subtract the area left of \(z = 1.23\), which from the table can be found to be approximately 0.8907. The difference gives \(P(1.23 鈮� z 鈮� 2.89) = 0.9981 - 0.8907 = 0.1074.\)
03

Find P(-2.24 鈮� z 鈮� -1.19)

Similarly, find the area to the left of \(z = -1.19\) from the Z-table, which is 0.1170, and subtract the area left of \(z=-2.24\), which is 0.0125. This gives the probability \(P(-2.24 鈮� z 鈮� -1.19) = 0.1170 - 0.0125 = 0.1045.\)
04

Find P(z < 2.02)

Finally, from the z-table, the area to the left of \(z = 2.02\) is found to be 0.9783. So, \(P(z < 2.02) = 0.9783.\)

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Key Concepts

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

Understanding Probability
Probability is a fundamental concept in statistics and everyday life. In the context of the standard normal distribution, probability represents the likelihood of a specific outcome within a continuous range of values.
  • For a standard normal distribution, probabilities are associated with z-scores, which reflect how many standard deviations a value is from the mean, usually zero.
  • Probabilities range from 0 to 1, where 0 means an impossible event and 1 signifies a certain event.
In simpler terms, the probability can tell us the chance of a continuous variable falling within a certain range. For example, in our exercises, when we calculate the probability like \( P(z < -1.31) \), we are determining how likely it is that a randomly selected value from the standard normal distribution is less than -1.31.
Navigating the Z-table
The z-table is a crucial tool for finding probabilities associated with the standard normal distribution. It helps us determine the area under the curve to the left of a given z-score.
  • A z-table usually shows the cumulative probability from the left tail of the distribution, up to a given z-score.
  • To find a probability between two z-scores, you can subtract the smaller z-score's cumulative probability from the larger one.
For instance, when you look up \( z = 1.23 \) and \( z = 2.89 \) in the table, you are finding the cumulative probabilities up to those points. Subtracting these gives you the probability of falling between \( z = 1.23 \) and \( z = 2.89 \), as seen in our exercise example.
Continuous Distribution Demystified
A continuous distribution like the standard normal distribution is one where data can take an infinite number of values within a certain range. Unlike discrete distributions, continuous distributions do not involve individual countable outcomes.
  • In a continuous distribution, probabilities represent areas under a curve, rather than distinct positions.
  • The total area under the curve of a standard normal distribution is 1, representing the sum of all possibilities.
This means that when we talk about the probability of \( z < 2.02 \), we are referring to the area under the curve from the far left up to a z-score of 2.02.
Decoding Z-scores
Z-scores are a way of standardizing data points within a normal distribution, allowing us to understand how far and in what direction a data point deviates from the mean.
  • A z-score of 0 indicates the value is exactly at the mean.
  • Positive z-scores show values above the mean, while negative z-scores indicate values below it.
  • The further a z-score is from zero, the more unusual or rare the data point is considered.
To use z-scores effectively, one needs to understand that they convert raw data into a standardized form, which can then be interpreted using the z-table to find probabilities, as demonstrated in our calculations for various ranges of z.

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Most popular questions from this chapter

According to an article on Yahoo.com on February 19,2012, the average salary of actuaries in the U.S. is \(\$ 98,620\) a year (http://education.yahoo.net/articles/careers_for_shy_people_2.htm?kid=1KWO3). Suppose that currently the distribution of annual salaries of all actuaries in the U.S. is approximately normal with a mean of \(\$ 98,620\) and a standard deviation of \(\$ 18,000\). How much would an actuary have to be paid in order to be in the highest-paid \(10 \%\) of all actuaries?

A machine at Kasem Steel Corporation makes iron rods that are supposed to be 50 inches long. However, the machine does not make all rods of exactly the same length. It is known that the probability distribution of the lengths of rods made on this machine is normal with a mean of 50 inches and a standard deviation of \(.06\) inch. The rods that are either shorter than \(49.85\) inches or longer than \(50.15\) inches are discarded. What percentage of the rods made on this machine are discarded?

Find the area under the standard normal curve a. between \(z=0\) and \(z=1.95\) b. between \(z=0\) and \(z=-2.05\) c. between \(z=1.15\) and \(z=2.37\) d. from \(z=-1.53\) to \(z=-2.88\) e. from \(z=-1.67\) to \(z=2.24\)

For a binomial probability distribution, \(n=25\) and \(p=.40\). a. Find the probability \(P(8 \leq x \leq 13)\) by using the table of binomial probabilities (Table I of Appendix C). b. Find the probability \(P(8 \leq x \leq 13)\) by using the normal distribution as an approximation to the binomial distribution. What is the difference between this approximation and the exact probability calculated in part a?

A study has shown that \(20 \%\) of all college textbooks have a price of \(\$ 184.52\) or higher. It is known that the standard deviation of the prices of all college textbooks is \(\$ 36.35 .\) Suppose the prices of all college textbooks have a normal distribution. What is the mean price of all college textbooks?
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