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Let \(x\) be a continuous random variable that is normally distributed with a mean of 25 and a standard deviation of 6 . Find the probability that \(x\) assumes a value a. between 29 and 36 b. between 22 and 35

Short Answer

Expert verified
The probability that \(x\) assumes a value between 29 and 36 is approximately 0.2178. The probability that \(x\) assumes a value between 22 and 35 is approximately 0.2610.

Step by step solution

01

Find the Z-Scores for Part (a)

Calculate the Z-scores for the limits of the given range (29 and 36) using the formula Z = (X – μ)/σ, where X is the value from the dataset, μ is the mean, and σ is the standard deviation. The mean μ is 25 and the standard deviation σ is 6. For X = 29, Z1 = (29 - 25) / 6 = 0.67 (rounded to two decimal places). For X = 36, Z2 = (36 - 25) / 6 = 1.83 (rounded to two decimal places).
02

Find the Probability for Part (a)

To find the probability that \(x\) falls between 29 and 36, we need to find the area under the standard normal curve between Z1 = 0.67 and Z2 = 1.83. Looking up Z1 and Z2 in the standard normal probability table, we find that the area corresponding to Z1 = 0.67 is 0.2514 and to Z2 = 1.83 is 0.0336. Subtracting these two areas, we find the area between Z1 and Z2 or our wanted probability is 0.2514 - 0.0336 = 0.2178.
03

Find the Z-Scores for Part (b)

We repeat the process from Step 1 for the values in range for part b, which are 22 and 35. For X = 22, Z1 = (22 - 25) / 6 = -0.5. For X = 35, Z2 = (35 - 25) / 6 = 1.67.
04

Find the Probability for Part (b)

We find the areas corresponding to the range's Z-scores as we did in Step 2. For Z1 = -0.5, the area is 0.3085 and for Z2 = 1.67, the area is 0.0475. Subtracting these areas we get the desired probability 0.3085 - 0.0475 = 0.2610.

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Key Concepts

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

Continuous Random Variable
A continuous random variable is a type of variable that can take infinite values within a given range. In the context of the exercise, the variable \(x\) is continuous, meaning it can assume any value between a lower and upper limit, unlike a discrete random variable that has countable outcomes.

This distinction is important because it determines how we calculate probabilities. For continuous random variables, probabilities are found for ranges of values rather than for specific, exact numbers. This probability is represented as the "area under the curve" of the distribution graph, also known as the normal distribution curve.

When dealing with continuous random variables, it is essential to use probability density functions. These functions help you find the likelihood of a variable falling within a particular range.
Z-Score
The Z-Score is a statistic that measures how many standard deviations an element is from the mean of the distribution. For example, in our exercise problem, Z-scores help determine how far the values 29, 36, 22, and 35 are from the mean of 25 in standard deviation units.

The Z-Score is calculated using the formula:
  • \( Z = \frac{X-\mu}{\sigma} \)
Where \(X\) is the value, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.

This measurement brings all data to the same scale, allowing comparison across different sets. In this exercise, calculating the Z-scores for given values transforms them into the standard normal distribution, a pivotal step for finding probabilities.
Standard Deviation
Standard deviation is a statistical measure that shows the amount of variance or dispersion in a set of values. A smaller standard deviation indicates that the data points tend to be closer to the mean, while a larger standard deviation signifies that they spread out over a broader range.

In our exercise, the standard deviation is given as 6, which means that the values of \(x\) typically deviate, or stray, 6 units from the average which is 25. This helps in understanding the spread of the distribution and in calculating the Z-Score more accurately.

Understanding standard deviation is crucial because it plays a fundamental role in the equation to transform values into Z-scores. It gives a standardized way of looking at how spread out or clustered together data is relative to the mean.
Probability
Probability in the context of a normal distribution and continuous random variables involves finding the area under the curve for a given range. In normal distribution, this curve is bell-shaped, symmetric about its mean.

To find the probability of \(x\) being between two points, you need the Z-scores for those points. These Z-scores are then looked up in a standard normal probability table, which provides the area under the curve to the corresponding Z-score.

For example, if you want to find the probability that \(x\) is between 29 and 36, and the Z-scores are 0.67 and 1.83, respectively, you consult the probability table to find the areas under the curve and subtract them. This difference further reveals the probability of the distributed values lying between those two bounds.

Probabilities from such calculations are fundamental in statistics as they provide insights into the likelihood of events occurring within specified ranges.

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Most popular questions from this chapter

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 receive free gifts?

Major League Baseball rules require that the balls used in baseball games must have circumferences between 9 and \(9.25\) inches. Suppose the balls produced by the factory that supplies balls to Major League Baseball have circumferences normally distributed with a mean of \(9.125\) inches and a standard deviation of \(.06\) inch. What percentage of these baseballs fail to meet the circumference requirement?

Two companies, A and B, drill wells in a rural area. Company A charges a flat fee of \(\$ 3500\) to drill a well regardless of its depth. Company B charges \(\$ 1000\) plus \(\$ 12\) per foot to drill a well. The depths of wells drilled in this area have a normal distribution with a mean of 250 feet and a standard deviation of 40 feet. a. What is the probability that Company B would charge more than Company A to drill a well? b. Find the mean amount charged by Company B to drill a well.

For a binomial probability distribution, \(n=120\) and \(p=.60\). Let \(x\) be the number of successes in 120 trials. a. Find the mean and standard deviation of this binomial distribution. b. Find \(P(x \leq 69)\) using the normal approximation. c. Find \(P(67 \leq x \leq 73)\) using the normal approximation.

The pucks used by the National Hockey League for ice hockey must weigh between \(5.5\) and \(6.0\) ounces. Suppose the weights of pucks produced at a factory are normally distributed with a mean of \(5.75\) ounces and a standard deviation of \(.11\) ounce. What percentage of the pucks produced at this factory cannot be used by the National Hockey League?

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