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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

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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

There is a difference in sports preferences between men and women, according to a recent survey. Among the 10 most popular sports, men include competition- type sports-pool and billiards, basketball, and softball-whereas women include aerobics, running, hiking, and calisthenics. However, the top recreational activity for men was still the relaxing sport of fishing, with \(41 \%\) of those surveyed indicating that they had fished during the year. Suppose 180 randomly selected men are asked whether they had fished in the past year. a. What is the probability that fewer than 50 had fished? b. What is the probability that between 50 and 75 had fished? c. If the 180 men selected for the interview were selected by the marketing department of a sporting goods company based on information obtained from their mailing lists, what would you conclude about the reliability of their survey results?

Consider a binomial random varible with \(n=25\) and \(p=.6 .\) Fill in the blanks below to find some probabilities using the normal approximation. a. Can we use the normal approximation? Calculate \(n p=\) _____ and \(n q=\) _____ b. Are \(n p\) and \(n q\) both greater than \(5 ?\) Yes ____ No ____ c. If the answer to part \(b\) is yes, calculate \(\mu=n p=\) ______ and \(\sigma=\sqrt{n p q}=\) ______ d. To find the probability of more than 9 successes, what values of \(x\) should be included? \(x=\) ________ e. To include the entire block of probability for the first value of \(x=\) ______, start at _______. f. Calculate \(z=\frac{x \pm .5-n p}{\sqrt{n p q}}=\) _______. g. Calculate \(P(x>9) \approx P(z>\)______) \(=1-\) _____ \(=\) ____.

Find a \(z_{0}\) such that \(P\left(-z_{0}

The Biology Data Book reports that the gestation time for human babies averages 278 days with a standard deviation of 12 days. \(^{8}\) Suppose that these gestation times are normally distributed. a. Find the upper and lower quartiles for the gestation times. b. Would it be unusual to deliver a baby after only 6 months of gestation? Explain.

Let \(x\) be a binomial random variable for \(n=25,\) \(p=.2\) a. Use Table 1 in Appendix I to calculate \(P(4 \leq x \leq 6)\). b. Find \(\mu\) and \(\sigma\) for the binomial probability distribution, and use the normal distribution to approximate the probability \(P(4 \leq x \leq 6)\). Note that this value is a good approximation to the exact value of \(P(4 \leq x \leq 6)\) even though \(n p=5\)
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