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Chapter 6: Problem 34

Let \(x\) be a continuous random variable that has a normal distribution with a mean of 40 and a standard deviation of 4 . Find the probability that \(x\) assumes a value a. between 29 and 35 b. from 34 to 50

Short Answer

Expert verified
The probability that \(x\) assumes a value between 29 and 35 is approximately 0.10267. The probability that \(x\) assumes a value between 34 and 50 is approximately 0.92698.

Step by step solution

01

Understand and impose the standardization

The goal is to compute the probability that \(x\) falls between two values. With \(x\) being a normally distributed variable with a mean of 40 and a standard deviation of 4. We standardize \(x\) in order to use the Z-table. The standardization of random variable \(x\) can be done by the following formula: \(z = \frac{x - \mu}{\sigma}\) where \(\mu\) is the mean and \(\sigma\) is the standard deviation.
02

Compute Z-scores for part a

Apply the Z-score formula for 29 and 35 respectively: For \(x = 29\), \(z_1 = \frac{29 - 40}{4} = -2.75\) For \(x = 35\), \(z_2 = \frac{35 - 40}{4} = -1.25\)
03

Find probabilities for part a

Consulting the Z-table, For \(z_1 = -2.75\), the corresponding probability is 0.00298 For \(z_2 = -1.25\), the corresponding probability is 0.10565 The probability that \(x\) is between 29 and 35 is therefore \(P(29 \leq x \leq 35) = P(-2.75 \leq Z \leq -1.25) = 0.10565 - 0.00298 = 0.10267\)
04

Compute Z-scores for part b

Apply the Z-score formula for 34 and 50 respectively: For \(x = 34\), \(z_3 = \frac{34 - 40}{4} = -1.5\) For \(x = 50\), \(z_4 = \frac{50 - 40}{4} = 2.5\)
05

Find probabilities for part b

Consulting the Z-table, For \(z_3 = -1.5\), the corresponding probability is 0.06681 For \(z_4 = 2.5\), the corresponding probability is 0.99379 The probability that \(x\) is between 34 and 50 is therefore \(P(34 \leq x \leq 50) = P(-1.5 \leq Z \leq 2.5) = 0.99379 - 0.06681 = 0.92698\)

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Key Concepts

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

Continuous Random Variable
In statistics, a continuous random variable is a type of variable that can take on an infinite number of values within a given range. Imagine measuring the height of a tree. The height could be 10.2 feet, 10.25 feet, or any possible value within that range. Continuous random variables are typically used to model situations involving measurements.
  • They can take any value within a specified range.
  • Often associated with measurements like time, weight, and temperature.
  • Pairs with probability density functions to determine the likelihood of a value occurring.
In our exercise, the variable \(x\) is continuous and follows a normal distribution, meaning it can hypothetically take any value. This makes calculating probabilities over a range of values an interesting task, as seen when determining the probabilities that \(x\) falls between certain numbers, such as 29 and 35, or 34 and 50.
Z-score
The Z-score is a statistical measure that tells us how many standard deviations a particular data point is from the mean of a data set. It's a way of standardizing or normalizing values, making it easier to compare them.
  • Shows how unusual or typical a data point is within a distribution.
  • A positive Z-score indicates a value above the mean, while a negative score is below.
  • Useful for comparing scores from different datasets.
In our exercise, the process of converting \(x\) values to Z-scores helps in finding probabilities using the Z-table. For example, calculating Z-scores for values 29 and 35 provides standardized scores, allowing us to easily read off probabilities from the Z-table.
Standardization
Standardization is the process of transforming data into a standard format. In the context of normal distribution, it involves converting a dataset with mean \(\mu\) and standard deviation \(\sigma\) into a standard normal distribution (mean of 0 and standard deviation of 1). This is done by transforming each value \(x\) in the dataset into a Z-score using the formula:
\[ z = \frac{x - \mu}{\sigma}\]
  • Transforms data to a common scale without changing the shape of the distribution.
  • Enables comparison across different datasets.
  • Facilitates use of statistical tools like the Z-table.
Through standardization, the exercise converts values 29 and 35 (or 34 and 50) into Z-scores. This makes it feasible to calculate probabilities using the Z-table, which is laid out for the standard normal distribution.
Z-table
A Z-table, or standard normal distribution table, is a type of statistical lookup table used to find the probability of a Z-score in a standard normal distribution. The Z-score determines the corresponding cumulative probability.
  • Most tables show the area to the left of a given Z-score.
  • Useful for finding probabilities and percentile ranks.
  • Widely used in hypothesis testing and confidence interval estimation.
In this exercise, the Z-table is used to find probabilities corresponding to Z-scores like -2.75 and -1.25 (for part a), and -1.5 and 2.5 (for part b). By consulting the table, we can determine how likely an event within a specific range is to occur, based on its position in the probability distribution.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. It quantifies how spread out numbers are in a dataset and is an important statistic in understanding data variability.
  • A small standard deviation indicates that the values tend to be close to the mean.
  • A large standard deviation suggests a wide range of values.
  • Crucial in calculating Z-scores and comparing different datasets.
In the original exercise, the standard deviation is noted as \(\sigma = 4\). This number plays a pivotal role in calculating Z-scores, as each data point is compared against the mean relative to this standard deviation. By knowing the standard deviation, predictions about how data is distributed around the mean can be made.

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Most popular questions from this chapter

What is the difference between the probability distribution of a discrete random variable and that of a continuous random variable? Explain.

One of the cars sold by Walt's car dealership is a very popular subcompact car called Rhino. The final sale price of the basic model of this car varies from customer to customer depending on the negotiating skills and persistence of the customer. Assume that these sale prices of this car are normally distributed with a mean of \(\$ 19,800\) and a standard deviation of \(\$ 350\). a. Dolores paid \(\$ 19,445\) for her Rhino. What percentage of Walt's customers paid less than Dolores for a Rhino? b. Cuthbert paid \(\$ 20,300\) for a Rhino. What percentage of Walt's customers paid more than Cuthbert for a Rhino?

The management of a supermarket wants to adopt a new promotional policy of giving a free gift to every customer who spends more than a certain amount per visit at this supermarket. The expectation of the management is that after this promotional policy is advertised, the expenditures for all customers at this supermarket will be normally distributed with a mean of \(\$ 95\) and a standard deviation of \(\$ 20\). If the management decides to give free gifts to all those customers who spend more than \(\$ 130\) at this supermarket during a visit, what percentage of the customers are expected to get free gifts?

Let \(x\) be a continuous random variable that follows a normal distribution with a mean of 550 and \(a\) standard deviation of 75 . a. Find the value of \(x\) so that the area under the normal curve to the left of \(x\) is 0250 . b. Find the value of \(x\) so that the area under the normal curve to the right of \(x\) is \(.9345\). c. Find the value of \(x\) so that the area under the normal curve to the right of \(x\) is approximately \(.0275\). d. Find the value of \(x\) so that the area under the normal curve to the left of \(x\) is approximately \(.9600\). e. Find the value of \(x\) so that the area under the normal curve between \(\mu\) and \(x\) is approximately \(.4700\) and the value of \(x\) is less than \(\mu\). f. Find the value of \(x\) so that the area under the normal curve between \(\mu\) and \(x\) is approximately \(.4100\) and the value of \(x\) is greater than \(\mu\).

How do the width and height of a normal distribution change when its mean remains the same but its standard deviation decreases?
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