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Let \(z\) denote a random variable that has a standard normal distribution. Determine each of the following probabilities: a. \(P(z<2.36)\) b. \(P(z \leq 2.36)\) c. \(P(z<-1.23)\) d. \(P(1.142)\) g. \(P(z \geq-3.38)\) h. \(P(z<4.98)\)

Short Answer

Expert verified
a. \(P(z<2.36)=0.9906\) b. \(P(z \leq 2.36)=0.9906\) c. \(P(z<-1.23)=0.1093\) d. \(P(1.142)=0.0228\) g. \(P(z \geq-3.38)=0.9996\) h. \(P(z<4.98) \approx 1\)

Step by step solution

01

a. P(Z < 2.36)

To find P(Z < 2.36), we must calculate the area under the PDF curve from negative infinity up to z = 2.36. For this, we use the Z-table or a calculator with the cumulative probability function of a standard normal distribution. The result we find for P(Z < 2.36) is 0.9906 (rounded to four decimal places).
02

b. P(Z ≤ 2.36)

A continuous distribution, like the standard normal distribution, assigns the same probability to any single value and a value with "≤". Hence, P(Z < 2.36) and P(Z ≤ 2.36) will have the same values that we calculated in the previous step. The result is P(Z ≤ 2.36) = 0.9906 (rounded to four decimal places).
03

c. P(Z < -1.23)

To find P(Z < -1.23), we must calculate the area under the PDF curve from negative infinity to z = -1.23. The result we find for P(Z < -1.23) is 0.1093 (rounded to four decimal places).
04

d. P(1.14 < Z < 3.35)

To find the probability that Z is between 1.14 and 3.35, we need to calculate the area under the PDF curve between these two values. We first find the cumulative probabilities for each value P(Z < 1.14) = 0.8729 and P(Z < 3.35) = 0.9996 (rounded to four decimal places). Then, we subtract the smaller cumulative probability from the larger one to get the area between 1.14 and 3.35. Therefore, P(1.14 < Z < 3.35) = 0.9996 - 0.8729 = 0.1267 (rounded to four decimal places).
05

e. P(-0.77 ≤ Z ≤ -0.55)

Since Z is continuous, P(-0.77 ≤ Z ≤ -0.55) = P(-0.77 < Z < -0.55). We find the cumulative probabilities for each value P(Z < -0.77) = 0.2206 and P(Z < -0.55) = 0.2912 (rounded to four decimal places). Then we subtract the smaller cumulative probability from the larger one to get the area between -0.77 and -0.55. Therefore, P(-0.77 < Z < -0.55) = 0.2912 - 0.2206 = 0.0706 (rounded to four decimal places).
06

f. P(Z > 2)

To find P(Z > 2), we must calculate the area under the PDF curve from z = 2 to positive infinity. We first find P(Z < 2) = 0.9772 (rounded to four decimal places). Since the total area under the PDF curve is equal to 1, P(Z > 2) = 1 - P(Z < 2) = 1 - 0.9772 = 0.0228.
07

g. P(Z ≥ -3.38)

Since Z is continuous, P(Z ≥ -3.38) = P(Z > -3.38). To find P(Z > -3.38), we must calculate the area under the PDF curve from z = -3.38 to positive infinity. We first find P(Z < -3.38) = 0.0004 (rounded to four decimal places). Then, P(Z > -3.38) = 1 - P(Z < -3.38) = 1 - 0.0004 = 0.9996.
08

h. P(Z < 4.98)

To find P(Z < 4.98), we must calculate the area under the PDF curve from negative infinity to z = 4.98. Since z = 4.98 is a very large value for a standard normal distribution, P(Z < 4.98) is approximately equal to 1. Using an online calculator or available statistical software, we can confirm that P(Z < 4.98) ≈ 1.

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Key Concepts

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

Z-table Usage
The Z-table, also known as the standard normal distribution table, is a valuable tool for understanding probabilities associated with a standard normal distribution. It represents cumulative probabilities of Z, a standard normal random variable, for different Z values. To use the Z-table, you first need to calculate your Z-value, which is the number of standard deviations a data point is from the mean. Once you have the Z-value, you look up the corresponding cumulative probability in the table. This cumulative probability is the probability that a standard normal random variable is less than or equal to that Z-value. For example, to find the probability that Z is less than 2.36, referred to as \( P(Z < 2.36) \), you would locate 2.36 in the Z-table to find the cumulative probability, which in this case is approximately 0.9906. Remember that Z-tables may present probabilities for the right tail (greater than Z) or the left tail (less than Z), so ensure you're using the table correctly based on your requirements.
Cumulative Probability
Cumulative probability refers to the probability that a random variable will take a value less than or equal to a certain number. In the context of the standard normal distribution, it's the total area under the probability density function (PDF) curve up to a specific Z-value. Essentially, it sums up the likelihood of a random variable falling within a given range, starting from the far left of the distribution (negative infinity) up to that point. Cumulative probability is essential in statistics because it allows us to determine the percentage of data that falls below a given Z-score. For example, the cumulative probability of \( P(Z \< -1.23) \) is 0.1093, meaning about 10.93% of the data in a standard normal distribution falls below a Z-score of -1.23. These probabilities help us understand the distribution's shape and make informed predictions or decisions based on the data.
Probability Density Function
The probability density function (PDF) is fundamental to understanding continuous probability distributions such as the standard normal distribution. To visualize it, think of the classic bell curve, where the highest point represents the mean. The curve shows the likelihood of different outcomes for our random variable Z. Even though the PDF itself doesn't provide exact probabilities (these are always zero for continuous variables), the area under the curve between two Z-values gives us the probability of Z falling within that range. For instance, the area under the curve from \( -\infty \) to \( Z = -1.23 \) represents \( P(Z \< -1.23) \). To calculate the probability of an interval, say \( P(1.14 \< Z \< 3.35) \), find the area under the PDF curve from 1.14 to 3.35. The PDF serves as the basis for defining the various cumulative probabilities we use when working with normal distributions.
Normal Distribution Calculations
Normal distribution calculations involve determining the probability of a random variable falling within a particular range. Most commonly, we calculate these probabilities with the standard normal distribution, where the mean is zero, and the standard deviation is one. The process typicall involves converting scores to Z-scores and then using the Z-table or cumulative density function to find probabilities. For example, calculating \( P(-0.77 \leq Z \leq-0.55) \) involves determining the cumulative probabilities at both Z-values and taking the difference to get the overall probability for the range. Understanding how to perform these calculations is essential for work in fields ranging from statistics to finance and psychology. It enables us to quantify the variability in data, and to make predictions about where future data points are likely to fall within a distribution.

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Most popular questions from this chapter

Suppose that your statistics professor tells you that the scores on a midterm exam were approximately normally distributed with a mean of 78 and a standard deviation of 7 . The top \(15 \%\) of all scores have been designated A's. Your score is \(89 .\) Did you earn an \(\mathrm{A}\) ? Explain.

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Let \(x\) denote the IQ of an individual selected at random from a certain population. The value of \(x\) must be a whole number. Suppose that the distribution of \(x\) can be approximated by a normal distribution with mean value 100 and standard deviation 15. Approximate the following probabilities. a. \(P(x=100)\) b. \(P(x \leq 110)\) c. \(P(x<110)\) (Hint: \(x<110\) is the same as \(x \leq 109 .\) ) d. \(P(75 \leq x \leq 125)\)

A company that manufactures mufflers for cars offers a lifetime warranty on its products, provided that ownership of the car does not change. Only \(20 \%\) of its mufflers are replaced under this warranty. a. In a random sample of 400 purchases, what is the approximate probability that between 75 and 100 (inclusive) mufflers are replaced under warranty? b. Among 400 randomly selected purchases, what is the approximate probability that at most 70 mufflers are replaced under warranty? c. If you were told that fewer than 50 among 400 randomly selected purchases were replaced under warranty, would you question the \(20 \%\) figure? Explain.

A restaurant has four bottles of a certain wine in stock. The wine steward does not know that two of these bottles (Bottles 1 and 2 ) are bad. Suppose that two bottles are ordered, and the wine steward selects two of the four bottles at random. Consider the random variable \(x=\) the number of good bottles among these two. a. When two bottles are selected at random, one possible outcome is (1,2) (Bottles 1 and 2 are selected) and another is (2,4). List all possible outcomes. b. What is the probability of each outcome in Part (a)? c. The value of \(x\) for the (1,2) outcome is 0 (neither selected bottle is good), and \(x=1\) for the outcome (2,4) . Determine the \(x\) value for each possible outcome. Then use the probabilities in Part (b) to determine the probability distribution of \(x\). (Hint: See Example \(6.5 .)\)

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