Let X be the random variable from normal distribution with mean (μ) = 60 and standard deviation (σ) = 12 Then, P ( 48 > X < 72 ) = ? Draw a bell curve.
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Let X be the random variable from
Then,
P ( 48 > X < 72 ) = ?
Draw a bell curve.
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- Let x be a random variable that represents white blood cell count per cubic milliliter of whole blood. Assume that x has a distribution that is approximately normal, with mean ? = 7450 and estimated standard deviation ? = 2350. A test result of x < 3500 is an indication of leukopenia. This indicates bone marrow depression that may be the result of a viral infection. (a) What is the probability that, on a single test, x is less than 3500? (Round your answer to four decimal places.)(b) Suppose a doctor uses the average x for two tests taken about a week apart. What can we say about the probability distribution of x? -The probability distribution of x is approximately normal with ?x = 7450 and ?x = 1661.70. -The probability distribution of x is approximately normal with ?x = 7450 and ?x = 2350. -The probability distribution of x is not normal. -The probability distribution of x is approximately normal with ?x = 7450 and ?x = 1175.00. What is the probability of x < 3500? (Round…The average number of miles (in thousands) that a car's tire will function before needing replacement is 66 and the standard deviation is 20. Suppose that 16 randomly selected tires are tested. Round all answers to 4 decimal places where possible and assume a normal distribution. What is the distribution of X? X ~ N(,) What is the distribution of ¯x¯? ¯x¯ ~ N(,) If a randomly selected individual tire is tested, find the probability that the number of miles (in thousands) before it will need replacement is between 70.3 and 77.6. For the 16 tires tested, find the probability that the average miles (in thousands) before need of replacement is between 70.3 and 77.6. For part d), is the assumption that the distribution is normal necessary? YesNoThe average number of miles (in thousands) that a car's tire will function before needing replacement is 66 and the standard deviation is 19. Suppose that 45 randomly selected tires are tested. Round all answers to 4 decimal places where possible and assume a normal distribution. What is the distribution of X? X ~ N( , ) What is the distribution of ¯x? x¯ ~ N( , ) If a randomly selected individual tire is tested, find the probability that the number of miles (in thousands) before it will need replacement is between 64.4 and 67. For the 45 tires tested, find the probability that the average miles (in thousands) before need of replacement is between 64.4 and 67. For part d), is the assumption that the distribution is normal necessary? NoYes
- You are to roll a fair die n=108 times, each time observing the number of dots appearing on the top side of the die. The number of dots showing on the topside of toss i is a random variable represented by Xi, i=1,2,⋯,108. (a) Consider the distribution of the random variable Xi. Find the mean and the standard deviation of the number of dots showing on the uppermost face of a single roll of this die.μXi= ____________ Enter your answer using all the decimals you can.σXi= ____________ Enter your answer using all the decimals you can. (b) What is the probability that the average topside on your tosses will be somewhere between 3.4 and 3.7? Enter your answer using all the decimals you can. _____________ (c) What is the probability that the average of your tosses is greater than 4.15? Enter your answer using all the decimals you can. ______________The percent of fat calories that a person in America consumes each day is normally distributed with a mean of about 36 and a standard deviation of 10. Suppose that one individual is randomly chosen. Let X = percent of fat calories. (a) Find the z-score corresponding to 40 percent of fat calories, rounded to 3 decimal places. (b) Find the probability that the percent of fat calories a person consumes is more than 40. (c) Shade the area corresponding to this probability in the graph below. (Hint: The x-axis is the z- score. Use your z-score from part (a), rounded to one decimal place). Shade: Left of a value Click and drag the arrows to adjust the values. -3 -2 -1 2 3 4 -1.5 (d) Find the maximum number for the lower quarter of percent of fat calories. Round your answer to 3 decimal places.Have American males (AMs) gotten heavier over the last 20 years? A random sample 77 AMs in 2019 had a mean weight of x = 189.030 pounds. A random sample 93 AMs in 1999 had a mean weight of y = 184.795 pounds. It is recognized that the true standard deviation of 2019 AMs weights is σx = 14.04 pounds while it is recognized that the true standard deviation of 1999 AMs weights is σy = 10.03 pounds. The true (unknown) mean of 2019 AMs weights is μx pounds, while the true (unknown) mean of 1999 AMs weights is μy pounds. Weights are known to be a normally distributed. In summary: Type Sample Size Sample Mean Standard Deviation 2019 Data (X) 77 189.030 14.04 1999 Data (Y) 93 184.795 10.03 g) What is the length of your 98% confidence interval for μx-μy? h) If we used this data to test H0:μx-μy=0 against the alternative Ha:μx-μy> 0 then what would the value of the calculated test statistic z have been?
- Let x be a random variable that represents white blood cell count per cubic milliliter of whole blood. Assume that x has a distribution that is approximately normal, with mean ? = 6050 and estimated standard deviation ? = 1500. A test result of x < 3500 is an indication of leukopenia. This indicates bone marrow depression that may be the result of a viral infection. (a) What is the probability that, on a single test, x is less than 3500? (Round your answer to four decimal places.) (b) Suppose a doctor uses the average x for two tests taken about a week apart. What can we say about the probability distribution of x? The probability distribution of x is not normal.The probability distribution of x is approximately normal with ?x = 6050 and ?x = 1500. The probability distribution of x is approximately normal with ?x = 6050 and ?x = 750.00.The probability distribution of x is approximately normal with ?x = 6050 and ?x = 1060.66. What is the probability of x < 3500? (Round your…The compressive strength of cement samples can be modeled by a normal distribution witha mean of (? + 6000) kilograms per square meter and a standard deviation of (? + 100)kilograms per square meter. Find the probability that a sample’s strength is between(? + 5800) kg/m2and (? + 5900) kg/m2 T = 28 ------------------------------------------------------------------------Assume that the amount of weight that male college students gain during their freshman year are normally distributed with a. Mean of u= 1.2 kg and a standard deviation of o= 5.1 kg A. If 1 male college student is randomly selected, Find the probability that he gains between 0 and 3 kg during freshman year B. If 25 male college students are randomly selected find the probability that their mean weight gain during freshman year is between 0 and 3 kg why can the normal distribution be used on pet b even though the sample size does not exceed 30?
- Have American males (AMs) gotten heavier over the last 20 years? A random sample 77 AMs in 2019 had a mean weight of x = 189.030 pounds. A random sample 93 AMs in 1999 had a mean weight of y = 184.795 pounds. It is recognized that the true standard deviation of 2019 AMs weights is σx = 14.04 pounds while it is recognized that the true standard deviation of 1999 AMs weights is σy = 10.03 pounds. The true (unknown) mean of 2019 AMs weights is μx pounds, while the true (unknown) mean of 1999 AMs weights is μy pounds. Weights are known to be a normally distributed. In summary: Type Sample Size Sample Mean Standard Deviation 2019 Data (X) 77 189.030 14.04 1999 Data (Y) 93 184.795 10.03 i) If we used this data to test H0:μx-μy=0 against the alternative Ha:μx-μy>0 then what would the p value have been? j)If we used this data to test H0:μx-μy=0 against the alternative Ha:μx-μy≠0 then what would the p value have been?A normal distribution has a mean of μ= 54 and a standard deviation of σ=6 What is the probability of selecting a sample of n = 4 scores with a mean more than M = 51?X is a normally distributed variable with mean = 60 and standard deviation = 10 %3D For a randomly selected score from this distribution what is the probability that X is less than 42
