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Given that \(z\) is a standard normal random variable, compute the following probabilities. a. \(\quad P(-1.98 \leq z \leq .49)\) b. \(\quad P(.52 \leq z \leq 1.22)\) c. \(\quad P(-1.75 \leq z \leq-1.04)\)

Short Answer

Expert verified
a) 0.6640, b) 0.1903, c) 0.1091

Step by step solution

01

Understand the Standard Normal Distribution

The standard normal distribution is a normal distribution with a mean of 0 and a standard deviation of 1. Probabilities are found using z-scores, which measure how many standard deviations an element is from the mean. We'll use a Standard Normal Distribution Table (z-table) to find probabilities.
02

Compute Probability for Part (a)

To find the probability \(P(-1.98 \leq z \leq 0.49)\), use the z-table. Find the probability for \(z = 0.49\), which is approximately 0.6879, and for \(z = -1.98\), which is approximately 0.0239. The probability is the difference between these values: \[ P(-1.98 \leq z \leq 0.49) = 0.6879 - 0.0239 = 0.6640 \]
03

Compute Probability for Part (b)

To find the probability \(P(0.52 \leq z \leq 1.22)\), use the z-table. Find the probability for \(z = 1.22\), which is approximately 0.8888, and for \(z = 0.52\), which is approximately 0.6985. The probability is the difference between these values: \[ P(0.52 \leq z \leq 1.22) = 0.8888 - 0.6985 = 0.1903 \]
04

Compute Probability for Part (c)

To find the probability \(P(-1.75 \leq z \leq -1.04)\), use the z-table. Find the probability for \(z = -1.04\), which is approximately 0.1492, and for \(z = -1.75\), which is approximately 0.0401. The probability is the difference between these values: \[ P(-1.75 \leq z \leq -1.04) = 0.1492 - 0.0401 = 0.1091 \]

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Key Concepts

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

Z-Scores
Z-scores are an essential concept when working with the standard normal distribution. A z-score shows how many standard deviations an individual data point or value is away from the mean of the dataset. In a standard normal distribution, the mean is 0 and the standard deviation is 1. Therefore, if you have a positive z-score, your data point is above the average. Conversely, a negative z-score indicates it is below the average. This allows you to compare different scores from various datasets on a common scale. To calculate a z-score, you use the formula: \[ z = \frac{(X - \mu)}{\sigma} \] Where:
  • \( X \) is the individual data point.
  • \( \mu \) is the mean of the dataset.
  • \( \sigma \) is the standard deviation of the dataset.
With standard normal distributions, working with z-scores is straightforward since the mean and standard deviation are known. This makes it easy to determine what percentage of data falls below, within, or above certain values.
Probability Calculation
Probability calculation in statistics refers to the process of finding the likelihood that a given event will occur within a particular distribution. In terms of the standard normal distribution, it usually entails calculating the probability that the random variable \(z\) falls within a certain range by using z-scores.For instance, if you want to calculate the probability \( P(a \leq z \leq b) \), where \(a\) and \(b\) are specific z-scores, you do this by finding the area under the standard normal curve between these scores. The process generally follows these steps:
  • Identify the z-scores that set the boundaries for your probability calculation. These correspond to points \(a\) and \(b\).
  • Use a z-table or calculator to find the probabilities associated with each z-score. This is the cumulative area under the distribution from the far left up to the z-score.
  • Subtract the smaller cumulative probability from the larger one to get the probability between the two z-scores.
For example, if you have probabilities 0.6879 for \( z = 0.49 \) and 0.0239 for \( z = -1.98 \), the calculation \[ 0.6879 - 0.0239 = 0.6640 \] gives you the probability \( P(-1.98 \leq z \leq 0.49) \). This represents the likelihood that a z-score will fall within this range.
Z-Table
A z-table, also known as the standard normal table, is a powerful tool used to find probabilities and percentiles for a given z-score in the standard normal distribution. The table contains the cumulative probability from the mean up to a given z-score, allowing us to easily determine the likelihood that a data point would fall within a certain range. When using a z-table, follow these simple steps:
  • Determine the z-score of interest, including whether it is positive or negative.
  • Locate the row that matches the first two digits (before the decimal) of the z-score.
  • In the same row, move across the columns to find the value in the column that matches the last digit of your z-score (usually presented up to two decimal places).
  • The number at the intersection is the cumulative probability. For example, if the z-score is 1.22, the z-table entry you find will show a probability of approximately 0.8888.
Remember that for negative z-scores, values will be below the mean and, as such, these probabilities are often less than 0.5. By subtracting cumulative probabilities for various z-scores, we can find the probability of a data point falling between two z-values, as shown in previous examples.

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Most popular questions from this chapter

Suppose we are interested in bidding on a piece of land and we know one other bidder is interested. \(^{1}\) The seller announced that the highest bid in excess of \(\$ 10,000\) will be accepted. Assume that the competitor's bid \(x\) is a random variable that is uniformly distributed between \(\$ 10,000\) and \(\$ 15,000\) a. Suppose you bid \(\$ 12,000\). What is the probability that your bid will be accepted? b. Suppose you bid \(\$ 14,000\). What is the probability that your bid will be accepted? c. What amount should you bid to maximize the probability that you get the property? d. Suppose you know someone who is willing to pay you \(\$ 16,000\) for the property. Would you consider bidding less than the amount in part (c)? Why or why not?

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