The weight of cereal boxes (y) follows a normal distributed with a variance of 0.1 squared ounces. If the average is 15.8 ounces, then P(y < 16) = If the average is 16.5 ounces, then P(y > 16.2)
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The weight of cereal boxes (y) follows a
- If the average is 15.8 ounces, then P(y < 16) =
- If the average is 16.5 ounces, then P(y > 16.2)
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- Assume the samples are random and independent, the populations are nomally distributed, and the population variances are equal. The table available below shows the prices (in dollars) for a sample of automobile batteries. The prices are classified according to battery type. At a = 0.10, is there enough evidence conclude that at least one mean battery price is different from the others? Complete parts (a) through (e) below. E Click the icon to view the battery cost data. (a) Let u1. P2. H3 represent the mean prices for the group size 35, 65, and 24/24F respectively. Identify the claim and state Ho and H. H Cost of batteries by type The claim is the V hypothesis. Group size 35 Group size 65 Group size 24/24F 101 111 121 124 D 146 173 182 278 124 140 141 89 (b) Find the critical value, Fo, and identify the rejection region. 90 79 84 The rejection region is F Fo, where Fo = (Round to two decimal places as needed.) (c) Find the test statistic F. Print Done F= (Round to two decimal places as…Our environment is very sensitive to the amount of ozone in the upper atmosphere. The level of ozone normally found is 7.97.9 parts/million (ppm). A researcher believes that the current ozone level is at an excess level. The mean of 2424 samples is 8.18.1 ppm with a variance of 0.250.25. Does the data support the claim at the 0.10.1 level? Assume the population distribution is approximately normal. Step 3 of 5: Specify if the test is one-tailed or two-tailed.Suppose that contamination particle size (in microm- eters) can be modeled as f(x)=2x³ for 1 < x. Determine the mean of X. What can you conclude about the variance of X?
- A company's single-serving cereal boxes advertise 1.63 ounces of cereal. In fact, the amount of cereal X in a randomly selected box can be modeled by a Normal distribution with a mean of 1.70 ounces and a standard deviation of 0.03 ounces. Let Y = the excess amount of cereal beyond what's advertised in a randomly selected box, measured in grams (1 ounce = 28.35 grams). A. Find the mean of Y B. Calculate and interpret the standard deviation of Y C. Find the probability of getting at least 1 gram more cereal than advertisedAssume the sample variances to be continuous measurements. Find the probability that a random sample of 31 observations, from a normal population with variance of 6, will have a sample variance between 3.33 and 9,99 Select one: Oa. 0.975 Ob. 0.011 Oc 0.985 Od. 0.965A firm produces calculators. It manager claims that its calculators are good, on average, for at least 58 months. A consumer protection agency tested 16 such calculators to check this claim. He found the mean life of these calculators to be 55 months with a variance of 9 months. Can you conclude that the claim of this company is true? Use a = 0.05. Assume that the life of such a calculator has an approximately normal distribution. According to this information the null and alternative hypotheses are: H0: μ ≤ 58 H1: μ > 58 Select one: True False
- The durability of the hacksaw is measured in a quality control laboratory. A standard saw needs 2500 cuts. The cutting average of 28 saws selected randomly from a company was found to be 2600 and the variance as 17500. I wonder if the company's production is above the standard? (α = 0.01)Our environment is very sensitive to the amount of ozone in the upper atmosphere. The level of ozone normally found is 7.9 parts/million (ppm). A researcher believes that the current ozone level is at an excess level. The mean of 24 samples is 8.1 ppm with a variance of 0.25. Does the data support the claim at the 0.1 level? Assume the population distribution is approximately normal. Step 5 of 5: Make the decision to reject or fail to reject the null hypothesis.Psi is a measure of compressive strength, or the ability of the material to carry loads and handle compression. The desired concrete psi rating used for sidewalks and residential driveways ranges from 2500psi to 3000psi obtained from mixing cement, stone, and sand in different ratios but with the same amount of water. The summary of the psi's of three (3) such concrete mixes made by three (3) different civil engineering students is given as follows: Mean vector: The variance-covariance matrix: Concrete mix Mean Concrete mix 1 Concrete mix 2 Concrete mix 3 Concrete mix 1 | Concrete mix 2 2700 18 12 2 3000 12 16 3 2400 Concrete mix 3 16 25 Let the random variable Y be the vector of the psi's of concrete mixes obtained by the students, i.e. Y1, denotes the psi of concrete mix made by student 1, Y2 is the psi of concrete mix made by student 2 and Y3 is the psi of concrete mix psi made by student 3. (a) Find (i) the multivariate probability distribution function (pdf) of Y. (ii) the…
- Total fat content (including saturated fat and trans fats) is a key characteristic of cheese. Assume that for a given type of cheese, total fat content (10-1 g/g of cheese) follows a normal distribution with a population mean of 2.85 (unit) and a population variance of 0.01 (unit2). Such an assumption is reasonable because fat content varies inside cheese and the “errors” around the mean total fat content (i.e. the value appearing on the etiquette) are distributed like a bell shape. a) What is the probability that the total fat content of this cheese exceeds 3 (unit)? b) What is the probability that the total fat content varies from 2.8 to 3 (unit)?A manufacturing company produces bearings. One line of bearings is specified to be 1.64 centimeters (cm) in diameter. A major customer requires that the variance of the bearings be no more than .001 cm². The producer is required to test the bearings before they are shipped, and so the diameters of 16 bearings are measured with a precise instrument, resulting in the following values. Assume bearing diameters are normally distributed. Use the data and a = 0.01 to test to determine whether the population of these bearings is to be rejected because of too high a variance. 1.68 1.62 1.63 1.70 1.66 1.63 1.65 1.71 1.64 1.69 1.57 1.64 1.59 1.66 1.63 1.65 Appendix A Statistical Tables (Round your answer to 2 decimal places, e.g. 15.25.) The value of the test statistic is and we reject the null hypothesis ⇒Sunscreens are products that protect the skin from skin cancer by preventing the sun's ultraviolet (UV) radiation from penetrating the skin. According to dermatologists, the recommended amount of sunscreen needed to cover the exposed areas of the body is at least an ounce. In a study conducted specifically for a group of women aged 21 to 30 years, it was found the amount of sunscreen (X) applied by the group follows the normal distribution with mean of 0.6 ounce and standard deviation of 0.25 ounce. The amount of sunscreen applied by women aged 21 to 30 years follows a bell-shaped distribution which is symmetric at X = _____.
