/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 94 Suppose that the distribution of... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Suppose that the distribution of the number of items \(x\) produced by an assembly line during an 8 -hr shift can be approximated by a normal distribution with mean value 150 and standard deviation \(10 .\) a. What is the probability that the number of items produced is at most 120 ? b. What is the probability that at least 125 items are produced? c. What is the probability that between 135 and 160 (inclusive) items are produced?

Short Answer

Expert verified
a) The probability that 120 or fewer items are produced is 0.13%. b) The probability that 125 or more items are produced is 99.38%. c) The probability that between 135 and 160 items are produced is 77.45%.

Step by step solution

01

Calculate Z-Score for 120

First you need to calculate the Z-score for 120. The formula for the Z score is \(Z = \frac{X - \mu}{\sigma}\), where \(X = 120\), \(\mu = 150\) and \(\sigma = 10\). After plugging in these values, you get \(Z = \frac{120 - 150}{10} = -3.0\)
02

Find Probability for At Most 120 Items

To find the probability that at most 120 items are produced, look up -3.0 in the standard normal distribution table. The value corresponding to -3.0 is 0.0013, so the probability is 0.0013 or 0.13%
03

Calculate Z-Score for 125

Next, calculate the Z-score for 125. Use the same formula as before. Plug in the values and you get \(Z = \frac{125 - 150}{10} = -2.5\)
04

Find Probability for At Least 125 Items

Look up -2.5 in the standard normal distribution table. The value is 0.0062. However, because you are looking for at least 125 items, you want to find the probability that the production is on the right side of -2.5. So, subtract 0.0062 from 1 to get the probability, which is \(1 - 0.0062 = 0.9938\) or 99.38%
05

Calculate Z-Scores for 135 and 160

Calculate the Z-Scores for 135 and 160 using the same formula. You get \(Z1 = \frac{135 - 150}{10} = -1.5\) and \(Z2 = \frac{160 - 150}{10} = 1.0\)
06

Find Probability for Between 135 and 160 Items

For the probability of production falling within 135 and 160 items, first look up the values for -1.5 and 1.0 in the standard normal distribution table. The corresponding values are 0.0668 and 0.8413. To find the probability, subtract the two values: \(0.8413 - 0.0668 = 0.7745\), or 77.45%

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Z-Score Calculation
If you've come across a normal distribution problem, one of the first steps you'll encounter is calculating the Z-score. The Z-score is a statistical measurement that describes a value's relationship to the mean of a group of values.

A Z-score can be calculated using the formula:
\[ Z = \frac{X - \mu}{\sigma} \]
where \(X\) is the value in question, \(\mu\) is the mean, and \(\sigma\) is the standard deviation. In a normal distribution, Z-scores indicate how many standard deviations an element is from the mean.

For example, if an assembly line produces, on average, \(150\) items with a standard deviation of \(10\), and you want to find the Z-score for \(120\) items, you would subtract the mean from \(120\) and then divide by the standard deviation:
\[ Z = \frac{120 - 150}{10} = -3.0 \]
This Z-score of \(-3.0\) signifies that \(120\) items are three standard deviations less than the average production.
Standard Normal Distribution Table
After calculating the Z-score, you’ll often need to determine the probability associated with that Z-score. This is where the standard normal distribution table comes into play. The table lists the probabilities for different Z-scores in a perfectly symmetrical bell-shaped distribution known as the standard normal distribution.

Since the table usually provides the probability that a value is less than the given Z-score, it is essential for finding cumulative probabilities. For example, let’s utilize the Z-score we previously calculated:
\[ Z = -3.0 \]
By referring to a standard normal distribution table, we find that the cumulative probability associated with \(Z = -3.0\) is approximately 0.0013, or 0.13%.

Understanding this table is critical when interpreting Z-scores. It translates the Z-score into a probability, enabling you to assess the likelihood of an event occurring within a normal distribution.
Probability in Normal Distribution
The beauty of a normal distribution is in its predictability; it's a tool that allows us to understand the probabilities of occurrences within a dataset. We can find out, for instance, the probability of an assembly line producing a specific number of items.

Probabilities in a normal distribution often involve finding areas under the distribution curve. To calculate these, we frequently use Z-scores and the standard normal distribution table.

Calculations for Different Scenarios

When determining the likelihood of producing at most or at least a certain number of items, we consider the cumulative probabilities:

- For 'at most', we look at the probability of being less than or equal to the Z-score.
- For 'at least', we take the complement of the 'at most' probability, which involves subtracting the table value from 1.

Additionally, when calculating the probability of a range, such as between 135 and 160 items, you would find the Z-scores for both values and then subtract the smaller Z-score's cumulative probability from the larger one.

These probabilities can help businesses and researchers make informed decisions based on anticipated outcomes.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Of all airline flight requests received by a certain discount ticket broker, \(70 \%\) are for domestic travel (D) and \(30 \%\) are for international flights (I). Let \(x\) be the number of requests among the next three requests received that are for domestic flights. Assuming independence of successive requests, determine the probability distribution of x. (Hint: One possible outcome is DID, with the probability \((.7)(.3)(.7)=.147 .\) )

Let \(x\) be the number of courses for which a randomly selected student at a certain university is registered. The probability distribution of \(x\) appears in the following table: $$ \begin{array}{lrrrrrrr} x & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ p(x) & .02 & .03 & .09 & .25 & .40 & .16 & .05 \end{array} $$ a. What is \(P(x=4)\) ? b. What is \(P(x \leq 4)\) ? c. What is the probability that the selected student is taking at most five courses? d. What is the probability that the selected student is taking at least five courses? more than five courses? e. Calculate \(P(3 \leq x \leq 6)\) and \(P(3

Accurate labeling of packaged meat is difficult because of weight decrease resulting from moisture loss (defined as a percentage of the package's original net weight). Suppose that moisture loss for a package of chicken breasts is normally distributed with mean value \(4.0 \%\) and standard deviation \(1.0 \% .\) (This model is suggested in the paper "Drained Weight Labeling for Meat and Poultry: An Economic Analysis of a Regulatory Proposal," Journal of Consumer Affairs [1980]: 307-325.) Let \(x\) denote the moisture loss for a randomly selected package. a. What is the probability that \(x\) is between \(3.0 \%\) and \(5.0 \%\) ? b. What is the probability that \(x\) is at most \(4.0 \%\) ? c. What is the probability that \(x\) is at least \(7.0 \%\) ? d. Find a number \(z^{*}\) such that \(90 \%\) of all packages have moisture losses below \(z^{*} \%\). e. What is the probability that moisture loss differs from the mean value by at least \(1 \%\) ?

A machine producing vitamin E capsules operates so that the actual amount of vitamin \(\mathrm{E}\) in each capsule is normally distributed with a mean of \(5 \mathrm{mg}\) and a standard deviation of \(0.05 \mathrm{mg} .\) What is the probability that a randomly selected capsule contains less than \(4.9 \mathrm{mg}\) of vita\(\min \mathrm{E} ?\) at least \(5.2 \mathrm{mg} ?\)

Bob and Lygia are going to play a series of Trivial Pursuit games. The first person to win four games will be declared the winner. Suppose that outcomes of successive games are independent and that the probability of Lygia winning any particular game is .6. Define a random variable \(x\) as the number of games played in the series. a. What is \(p(4)\) ? (Hint: Either Bob or Lygia could win four straight games.) b. What is \(p(5) ?\) (Hint: For Lygia to win in exactly five games, what has to happen in the first four games and in Game \(5 ?\) ) c. Determine the probability distribution of \(x\). d. How many games can you expect the series to last?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.