The number of flaws per square yard in a type of carpet material varies with mean 1.6flaws per square yard and standard deviation flaws per square yard.

a. Without doing any calculations, explain which event is more likely:

  • randomly selecting a1square yard of material and finding 2 or more flaws
  • randomly selecting 50square yards of material and finding an average of 2or more flaws

b. Explain why you cannot use a Normal distribution to calculate the probability of the first event in part (a).

c. Calculate the probability of the second event in part (a).

Short Answer

Expert verified

a. Choosing one square yard of material at random and discovering two or more defects

b. The population distribution's shape is uncertain, and the sample size is limited.

c. The resultant probability of part(a) is 0.81%

Step by step solution

01

Part (a) Step 1: Given Information

The number of flaws per square yard in a type of carpet material varies with mean 1.6 flaws per square yard and standard deviation 1.2 flaws per square yard.

02

Part (a) Step 2: According to the given question

The population standard deviation divided by the square root of sample size equals the standard deviation of the sampling distribution of the sample mean.

σx¯=σn

As a result, the standard deviation lowers as the sample size grows, and the data values become closer to the predicted value as the sample size grows.

As a result, when the sample size is bigger, you are less likely to find a mean of two or more flaws, and when the sample size is less, you are more likely to find a mean of two or more flaws.

03

Part (b) Step 1: Given Information

The number of flaws per square yard in a type of carpet material varies with mean1.6 flaws per square yard and standard deviation 1.2 flaws per square yard.

04

Part (b) Step 2: According to the given question

Consider ,n=1

The center limit theorem states that if the sample size is more than 30, the sampling distribution of the sample mean x¯is approximately normal.

The central limit theorem cannot be applied since the sample size of 1 is less than 30. The sample mean sampling distribution has the same shape as the population distribution in this example.

Although the population distribution is unknown, the form of the sampling distribution of the sample mean is also unknown, which means that the probability cannot be calculated.

05

Part (c) Step 1: Given Information

The number of flaws per square yard in a type of carpet material varies with mean 1.6 flaws per square yard and standard deviation1.2flaws per square yard.

06

Part (c) Step 2: According to the given question

Consider that ,

μ=16σ=1.2n=50x¯=2

The following concept was used:

z=xμx¯σx¯
z=xμx¯σx¯

z-score is localid="1657630095652" z=x-μx¯σx¯=x¯-μσ/n=2-1.61.250=2.36

The z-two score's found expressions must then be equal:

localid="1657629821949" P(Z<2.36)is given in the row starting with.01and in the column starting with .06of the standard normal probability table

localid="1657629877918" P(X¯2)=P(Z>2.36)=1P(Z<2.36)=10.9209=0.0081=0.81%

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 Vaia!

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

Bottling cola A bottling company uses a filling machine to fill plastic bottles with cola. The bottles are supposed to contain 300 milliliters (ml). In fact, the contents vary according to a Normal distribution with mean μ=298mland standard deviation σ=3ml.

a. What is the probability that a randomly selected bottle contains less than 295ml?

b. What is the probability that the mean contents of six randomly selected bottles is less than 295ml?

Suppose we select an SRS of size n=100n=100from a large population having proportion p of successes. Let pp^be the proportion of successes in the sample. For which value of p would it be safe to use the Normal approximation to the sampling distribution of pp^?

a. 0.01

b. 0.09

c. 0.85

d. 0.975

e. 0.999

The probability distribution for the number of heads in four tosses of a coin is given by

Number of heads
01234
Probability
0.06250.2500
0.3750
0.2500
0.0625

The probability of getting at least one tail in four tosses of a coin is

a. 0.2500

b. 0.3125

c. 0.6875

d. 0.9375

e. 0.0625

A newborn baby has extremely low birth weight (ELBW) if it weighs less than 1000 grams. A study of the health of such children in later years examined a random sample of 219 children. Their mean weight at birth was x-x¯=810grams. This sample mean is an unbiased estimator of the mean weight μ in the population of all ELBW babies, which means that

a. in all possible samples of size 219 from this population, the mean of the values of x-x¯will equal 810 .

b. in all possible samples of size 219 from this population, the mean of the values of x-x¯will equal μ

c. as we take larger and larger samples from this population, x-x¯will get closer and closer to μ.

d. in all possible samples of size 219 from this population, the values of x-x¯will have a distribution that is close to Normal.

e. the person measuring the children's weights does so without any error.

Detecting gypsy moths The gypsy moth is a serious threat to oak and aspen trees. A state agriculture department places traps throughout the state to detect the moths. Each month, an SRS of 50 traps is inspected, the number of moths in each trap is recorded, and the mean number of moths is calculated. Based on years of data, the distribution of moth counts is discrete and strongly skewed with a mean of 0.5 and a standard deviation of 0.7.

a. Explain why it is reasonable to use a Normal distribution to approximate the sampling distribution of x-x¯for SRSs of size 50 .

b. Estimate the probability that the mean number of moths in a sample of size 50 is greater than or equal to 0.6.

c. In a recent month, the mean number of moths in an SRS of size 50 was x-=0.6. x¯=0.6. Based on this result, is there convincing evidence that the moth population is getting larger in this state? Explain your reasoning.

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.

Sign-up for free