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Chapter 6: Problem 56

a. Find the probability that \(z\) is greater than \(-.75 .\) b. Find the probability that \(z\) is less than 1.35 .

Short Answer

Expert verified
Answer: (a) The probability that z is greater than -0.75 is 0.2266. (b) The probability that z is less than 1.35 is 0.9115.

Step by step solution

01

Finding the probability that z is greater than -0.75

In this step, we will find the probability that the z-score is greater than -0.75. Since the standard normal distribution is symmetrical about the mean, the area to the right of -0.75 is equal to the area to the left of 0.75. First, we need to check the standard normal distribution table for the probability corresponding to z = 0.75. The table value for z = 0.75 is 0.7734. This means that the area to the left of z = 0.75 is 0.7734. Since the total area under the curve is equal to 1, we can calculate the area to the right of z = -0.75 as follows: P(z > -0.75) = 1 - P(z < 0.75) = 1 - 0.7734 = 0.2266 So, the probability that z is greater than -0.75 is 0.2266.
02

Finding the probability that z is less than 1.35

In this step, we will find the probability that the z-score is less than 1.35. We need to check the standard normal distribution table for the probability corresponding to z = 1.35, which is given by the area to the left of z = 1.35. The table value for z = 1.35 is 0.9115. This means that the area to the left of z = 1.35 is 0.9115. So, the probability that z is less than 1.35 is 0.9115. Now, we have the answers to both tasks: a. The probability that z is greater than -0.75 is 0.2266. b. The probability that z is less than 1.35 is 0.9115.

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Key Concepts

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

Standard Normal Distribution
In probability theory, the concept of the standard normal distribution is fundamental. It is a type of normal distribution that is fully characterized by having a mean (\( \mu \)) of 0 and a standard deviation (\( \sigma \)) of 1. This distribution is denoted as\( N(0,1) \). The standard normal distribution is represented by a bell-shaped curve that is symmetrical around its mean.

Key features of the standard normal distribution include:
  • Symmetrical shape: The curve is identical on both sides of the mean.
  • Total area under the curve: It sums up to 1, covering the probability of all possible outcomes.
  • 68-95-99.7 rule: About 68% of values lie within one standard deviation from the mean, 95% within two, and 99.7% within three.
Transforming any normal distribution to a standard normal distribution simplifies calculations, especially when utilizing z-scores and probability tables.
Z-Score
A z-score is an essential statistical measurement that expresses the number of standard deviations a data point is from the mean of a distribution. It is calculated using the formula:\[z = \frac{(X - \mu)}{\sigma}\]Here, \( X \) represents the value for which the z-score is being calculated, \( \mu \) is the mean of the distribution, and \( \sigma \) is the standard deviation.

Z-scores help in evaluating how extreme a point is within the crux of a distribution. If a z-score is:
  • Positive: the value is above the mean.
  • Negative: the value is below the mean.
  • Zero: the value is equal to the mean.
By converting raw scores to z-scores, we can easily compare data from different distributions and determine probabilities using the standard normal distribution.
Probability Table
A probability, or standard normal distribution, table is a handy mathematical tool used to find the probability of a z-score in a standard normal distribution. It provides the cumulative probability that a normally distributed random variable is less than or equal to a given z-score.

Using the probability table involves the following:
  • Identifying the z-score for which you want to find the probability.
  • Locating the row corresponding to the integer and first decimal place of the z-score.
  • Finding the column corresponding to the second decimal place of the z-score.
For example, if you are looking for the probability that a z-score is less than 1.35, you would find 0.9115 in the table. The table assumes values to the left of the curve. For values to the right, you subtract the table value from 1, accommodating the entire distribution's total area of 1.

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Most popular questions from this chapter

Cerebral blood flow (CBF) in the brains of healthy people is normally distributed with a mean of 74 and a standard deviation of 16 a. What proportion of healthy people will have CBF readings between 60 and \(80 ?\) b. What proportion of healthy people will have CBF readings above \(100 ?\) c. If a person has a CBF reading below \(40,\) he is classified as at risk for a stroke. What proportion of healthy people will mistakenly be diagnosed as "at risk"?

For a car traveling 30 miles per hour (mph), the distance required to brake to a stop is normally distributed with a mean of 50 feet and a standard deviation of 8 feet. Suppose you are traveling \(30 \mathrm{mph}\) in a residential area and a car moves abruptly into your path at a distance of 60 feet. a. If you apply your brakes, what is the probability that you will brake to a stop within 40 feet or less? Within 50 feet or less? b. If the only way to avoid a collision is to brake to a stop, what is the probability that you will avoid the collision?

Calculate the area under the standard normal curve to the left of these values: a. \(z=1.6\) b. \(z=1.83\) c. \(z=.90\) d. \(z=4.18\)

Students very often ask their professors whether they will be "curving the grades." The traditional interpretation of "curving grades" required that the grades have a normal distribution, and that the grades will be assigned in these proportions: $$ \begin{array}{l|lllll} \text { Letter Grade } & \mathrm{A} & \mathrm{B} & \mathrm{C} & \mathrm{D} & \mathrm{F} \\ \hline \text { Proportion of Students } & 10 \% & 20 \% & 40 \% & 20 \% & 10 \% \end{array} $$ a. If the average "C" grade is centered at the average grade for all students, and if we assume that the grades are normally distributed, how many standard deviations on either side of the mean will constitute the "C" grades? b. How many deviations on either side of the mean will be the cutoff points for the "B" and "D" grades?

The life span of oil-drilling bits depends on the types of rock and soil that the drill encounters, but it is estimated that the mean length of life is 75 hours. Suppose an oil exploration company purchases drill bits that have a life span that is approximately normally distributed with a mean equal to 75 hours and a standard deviation equal to 12 hours. a. What proportion of the company's drill bits will fail before 60 hours of use? b. What proportion will last at least 60 hours? c. What proportion will have to be replaced after more than 90 hours of use?
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