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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

The accompanying data on \(x=\) student-teacher ratio is for a random sample of 20 high schools in Maine selected from a population of 85 high schools. The data are consistent with summary values for the state of Maine that appeared in an article in the Bangor Daily News (September \(22,2016,\) bangordailynews.com/2016/09/22/mainefocus/we-discovered-a-surprise-when-we- looked-deeper-into-our-survey-of-maine-principals/?ref=morelnmidcoast, retrieved May 2, 2017). The corresponding normal scores are also shown. $$ \begin{array}{|cc|} \hline \text { Student-Teacher Ratio }(x) & \text { Normal Score } \\ \hline 9.0 & -1.868 \\ 10.0 & -1.403 \\ 11.0 & -1.128 \\ 11.2 & -0.919 \\ 11.6 & -0.744 \\ 11.7 & -0.589 \\ 11.8 & -0.448 \\ 11.9 & -0.315 \\ 12.0 & -0.187 \\ 12.1 & -0.062 \\ 12.5 & 0.062 \\ 12.6 & 0.187 \\ 13.0 & 0.315 \\ 13.2 & 0.448 \\ 13.6 & 0.589 \\ 13.7 & 0.744 \\ 14.0 & 0.919 \\ 14.5 & 1.128 \\ 14.9 & 1.403 \\ 15.0 & 1.868 \\ \hline \end{array} $$ a. Construct a normal probability plot. b. Calculate the correlation coefficient for the (normal score, \(x\) ) pairs. Compare this value to the appropriate critical \(r\) value from Table 6.2 to determine if it is reasonable to think that the distribution of student-teacher ratios for high schools in Maine is approximately normal.

Twenty-five percent of the customers of a grocery store use an express checkout. Consider five randomly selected customers, and let \(x\) denote the number among the five who use the express checkout. a. Calculate \(p(2),\) that is, \(P(x=2)\). b. Calculate \(P(x \leq 1)\). c. Calculate \(P(x \geq 2)\). (Hint: Make use of your answer to Part (b).) d. Calculate \(P(x \neq 2)\).

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A machine that cuts corks for wine bottles operates in such a way that the distribution of the diameter of the corks produced is well approximated by a normal distribution with mean \(3 \mathrm{~cm}\) and standard deviation \(0.1 \mathrm{~cm} .\) The specifications call for corks with diameters between 2.9 and \(3.1 \mathrm{~cm}\). A cork not meeting the specifications is considered defective. (A cork that is too small leaks and causes the wine to deteriorate; a cork that is too large doesn't fit in the bottle.) What proportion of corks produced by this machine are defective?

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