paint manufacturer produces cans of paint of nominal volume 2 litres. In fact the volume of paint in a can is a Normal random variable with mean 2.02 litres and standard deviation 0.02 litres. A single can is examined; the probability that it contains less than 2 litres is (3 d.p.)?
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A paint manufacturer produces cans of paint of nominal volume 2 litres.
In fact the volume of paint in a can is a Normal random variable with
A single can is examined; the probability that it contains less than 2 litres is (3 d.p.)?
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- A machine produces glass with a nominal thickness of 4mm. In fact the thickness is a Normal random variable with mean 4.168mm and standard deviation 0.12mm. The thickness in mm which 5% of panes exceed is (2 d.p.)?Gentle Ben is a Morgan horse at a Colorado dude ranch. Over the past 8 weeks, a veterinarian took the following glucose readings from this horse (in mg/100 ml). 94 87 83 104 98 108 84 89 The sample mean is x = 93.4. Let x be a random variable representing glucose readings taken from Gentle Ben. We may assume that x has a normal distribution, and we know from past experience that o = 12.5. The mean glucose level for horses should be u = 85 mg/100 ml.t Do these data indicate that Gentle Ben has an overall average glucose level higher than 85? Use a = 0.05. (a) What is the level of significance? State the null and alternate hypotheses. Will you use a left-tailed, right-tailed, or two-tailed test? O Ho: H = 85; H,: u + 85; two-tailed O Ho: H = 85; H: µ > 85; right-tailed O Ho: H > 85; H,: µ = 85; right-tailed O Ho: H = 85; H,: µ < 85; left-tailed (b) What sampling distribution will you use? Explain the rationale for your choice of sampling distribution. O The Student's t, since we assume…The average cost per household of owning a brand new car is $5,500. Suppose that we randomly selected 45 households, determine the probability that the sample mean for these 45 households is between $5,250 and 6,000. Assume that the variable is normally distributed and the standard deviation is $1,550. SD of the sample mean (input 4 decimal places)= type your answer... type your answer... z-scores (input 2 decimal places) type your answer... Final answer/Probability (input 2 decimal places type your answer... %
- In a study of life expectancy in a certain region, the mean age at death was 70 years, and the standard deviation was 5 years. The life expectancy (years) is assumed to follow a skewed distribution. Suppose that a random sample of 45 people from this region is selected. The mean life expectancy (years) of the sample is considered a random variable. What is its standard error? 0.5976 O 2.2361 O 5.2341 O 0.1111MN.27.Given that Z is a standard normal random variable, what is the value of Z if the area to the right of Z is 0.7054? A.-0.54 B.1.54 C.0.54 D.-1.54
- Coal is carried from a mine in west virginia to a power plant in new york in hopper cars on a long train. The automatic hopper car loader is set to put 77 tons of coal into each car. The actual weights of coal loaded into each care are normally distributed with mean u= 77 tons and standard deviation 0= 1.1 ton. A. What is the probability that one car chosen at random will have less than 76.5 tons of coal? ( Round your answer to four decimal places.) B. What is the probability that 29 cars chosen at random will have a mean load weight x of less than 76.5 tons of coal? ( Round your answer to four decimal places) C. Suppose the weight of coal in one car was less than 76.5 tons. Would that fact make you suspect that the loader had slipped out of adjustment. Suppose the weight of coal in 29 cars selected at random had an average x of less than 76.5 tons. Would that fact make you suspect that the loader had slipped out of adjustment? Why?The general philosophy of standardizing is that standardizing a random variable involves a. Subtracting the main in the batting by the standard deviation b. Dividing the standard deviation of the total of successes by the total of trials c. Adding and subtracting standard deviations in each direction d. Subtracting the area in the right tail from 1 since the Z chart gives only the area "to the left"The mean height of 15-year-old boys (in cm) is 175 and the variance is 64. For girls, the mean is 165 and the variance is 64. Assuming that the two populations are normally distributed. If eight boys and eight girls were sampled, then the probability that the mean height of the sample of boys will exceed the mean height of the sample of girls by 1 or more is: O 0.9938 0.9878 0.0062 O 0.9972
- The distribution of weekly salaries at a large company has a mean of $1000 and a standard deviation of $350. What is the probability that the mean of a random sample of weekly salaries of 50 employees will be at most $900? 0.0217 O 0.9783 O 0.3438 O Cannot be determined, because the distribution of the population is not normal.Two normally distributed populations with H4 =µ, and both variances being 100 are given. Random samples of size 25 are taken from each population with respective sample means 104 and 100. What is the probability of observing a difference of at least 4 units between the first and second sample means? Select one: O a.None b.0.1271 c.0.8729 d.0.9207 e.0.0793Indrilling a vertical wellbore, the spacing between major fractures are recorded. It has been found that this spacing is having a mean of 180o ft and a variance of 3,240,000 ft In the first 2000 ft drilling, there was no present of major fracture in the well. Determine the probability that there is a major fracture at the depth of 2000-3500 ft. Select one: a.0.57 b.o.43 C. O.19 do 82