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For the standard normal distribution, what is the area within \(2.5\) standard deviations of the mean?

Short Answer

Expert verified
The area under the curve of the standard normal distribution within 2.5 standard deviations of the mean is approximately 0.9876 or 98.76%.

Step by step solution

01

Understand the Standard Normal Distribution

In a standard normal distribution, mean is 0. The problem asks for the area within 2.5 standard deviations of the mean. This is equivalent to finding the area between \(Z = -2.5\) and \(Z = 2.5\). The Z-score tells you how many standard deviations you are away from the mean.
02

Look up the area from Z-table

Look up the value from a standard normal distribution table for \(Z = 2.5\). This gives the cumulative probability from the left side of the curve up to 2.5 standard deviations, which is approx. 0.9938 or 99.38%.
03

Do the calculation

Subtract the cumulative probability of \(Z = -2.5\) from the cumulative probability of \(Z = 2.5\). Because of symmetry, the cumulative probability of \(Z = -2.5\) is \(1 - 0.9938 = 0.0062\). Therefore, subtract 0.0062 from 0.9938 to get the final result.

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Key Concepts

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

Z-score
The Z-score is a statistical measurement that tells us how many standard deviations a data point is from the mean of the dataset. In a standard normal distribution, the mean is 0 and the standard deviation is 1, which simplifies calculations. A Z-score of 0 indicates the data point is exactly at the mean, while positive or negative Z-scores signify points above or below the mean, respectively. It's a way to standardize different data points, allowing for comparison across different datasets with varying means and standard deviations. For example, a Z-score of 2.5 means the data point is 2.5 standard deviations above the mean. Similarly, a Z-score of -2.5 would mean it's 2.5 standard deviations below the mean.

The Z-score is calculated using the formula:
\(Z = \frac{(X - \mu)}{\sigma}\)
where \(X\) is the value of the element, \(\mu\) is the mean of the data, and \(\sigma\) is the standard deviation. This formula allows you to convert any data point into the common scale of the standard normal distribution, facilitating comparisons and further statistical analysis.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. A low standard deviation indicates that values tend to be close to the mean of the set, while a high standard deviation indicates that the values are spread out over a wider range. It's a critical metric in statistics to understand how data is spread out.

In mathematical terms, standard deviation is the square root of the variance. The formula for standard deviation is:
\( \sigma = \sqrt{\frac{1}{N} \sum_{i=1}^{N}(X_i - \mu)^2} \)
where \(N\) is the number of data points, \(X_i\) represents each data point, and \(\mu\) is the mean of the data.

Understanding standard deviation is crucial when dealing with a normal distribution because it quantifies the amount of spread in the data. In the context of the standard normal distribution, the standard deviation is always 1. This simplification makes it easy to compare different datasets or to find probabilities using the Z-score.
Cumulative Probability
Cumulative probability represents the likelihood that a random variable is less than or equal to a given value. In the context of a standard normal distribution, it allows us to determine the probability of a Z-score falling within a certain range.

When we look up a Z-score in a standard normal distribution table (often called a Z-table), we find the cumulative probability up to that Z-score. For example, a cumulative probability of 0.9938 for a Z-score of 2.5 means roughly 99.38% of the data lies below that point.

The concept of cumulative probability is vital to understanding how data is distributed relative to the mean. In our example, to find the area within 2.5 standard deviations of the mean, you calculate the cumulative probability from \(Z = -2.5\) to \(Z = 2.5\). This involves subtracting the cumulative probability of \(Z = -2.5\) from that of \(Z = 2.5\). The result indicates the percentage of data within this range, which is crucial in making data-driven decisions and predictions.

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Most popular questions from this chapter

Johnson Electronics makes calculators. Consumer satisfaction is one of the top priorities of the company's management. The company guarantees the refund of money or a replacement for any calculator that malfunctions within two years from the date of purchase. It is known from past data that despite all efforts, \(5 \%\) of the calculators manufactured by this company malfunction within a 2 -year period. The company recently mailed 500 such calculators to its customers. a. Find the probability that exactly 29 of the 500 calculators will be returned for refund or replacement within a 2-year period. b. What is the probability that 27 or more of the 500 calculators will be returned for refund or replacement within a 2 -year period? c. What is the probability that 15 to 22 of the 500 calculators will be returned for refund or replacement within a 2 -year period?

Obtain the area under the standard normal curve a. to the right of \(z=1.43\) b. to the left of \(z=-1.65\) c. to the right of \(z=-.65\) d. to the left of \(z=.89\)

The management at Ohio National Bank does not want its customers to wait in line for service for too long. The manager of a branch of this bank estimated that the customers currently have to wait an average of 8 minutes for service. Assume that the waiting times for all customers at this branch have a normal distribution with a mean of 8 minutes and a standard deviation of 2 minutes a. Find the probability that a randomly selected customer will have to wait for less than 3 minutes. b. What percentage of the customers have to wait for 10 to 13 minutes? c. What percentage of the customers have to wait for 6 to 12 minutes? d. Is it possible that a customer may have to wait longer than 16 minutes for service? Explain.

According to the records of an electric company serving the Boston area, the mean electricity consumption during winter for all households is 1650 kilowatt-hours per month. Assume that the monthly electric consumptions during winter by all households in this area have a normal distribution with a mean of 1650 kilowatt-hours and a standard deviation of 320 kilowatt-hours. The company sent a notice to Bill Johnson informing him that about \(90 \%\) of the households use less electricity per month than he does. What. is Bill Johnson's monthly electricity consumption?

quarter), Britons spend an a… # According to a 2004 survey by the telecommunications division of British Gas (Source: http://www. literacytrust.org.uk/Database/texting.html#quarter), Britons spend an average of 225 minutes per day communicating electronically (on a fixed landline phone, on a mobile phone, by emailing, by texting, and so on). Assume that currently such times for all Britons are normally distributed with a mean of 225 minutes per day and a standard deviation of 62 minutes per day. What percentage of Britons communicate electronically for a. less than 60 minutes per day b. more than 360 minutes per day c. between 120 and 180 minutes per day d. between 240 and 300 minutes per day?

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