SITUATION 8 (a) The elongation of a steel bar under a particular load has been established to be normally distributed with mean of 0.05 inch and standard deviation of 0.01 inch. (a.1) Find the probability that the elongation of a steel bar is above 0.1 inch. (a.2) Find the probability that the elongation of a steel bar is between 0.025 and 0.065 inch.
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- If the variable Z is normally distributed with mean equal to 0 and standard deviation equal to 1. The probability that the variable takes a value less than -0.9 or greater than 0.5 is: Select one: P(Z < -0.9 or Z > 0.5) = 0.5074 P(Z < -0.9 or Z > 0.5) = 0.1915 P(Z < -0.9 or Z > 0.5) = 0.3159 P(Z < -0.9 or Z > 0.5) = 0.4926Porosity measurements in an oil field show a normal distribution witha mean of 5% and a standard deviation of 1%. Determine the probability of porosity to be more than 6%.Assume that women's heights are normally distributed with a mean given by μ=63.5 in,and a standard deviation given by σ=2.3 in. (a) If 1 woman is randomly selected, find the probability that her height is less than 64 in. (b) If 48 women are randomly selected, find the probability that they have a mean height less than 64 in. (a)The probability is approximately ?
- Express the k th percentile, Pk , of a normally distributed variable in terms of its mean, µ, and standard deviation, σ.nutritionist claims that the mean tuna consumption by a person is 3.4 pounds per year. A sample of 90 people shows that the mean tuna consumption by a person is 3.3 pounds per year. Assume the population standard deviation is 1.02 pounds. At α=0.03, can you reject the claim? (a) Identify the null hypothesis and alternative hypothesis. A. H0: μ=3.4 Ha: μ≠3.4 Your answer is correct. B. H0: μ>3.4 Ha: μ≤3.4 C. H0: μ≤3.3 Ha: μ>3.3 D. H0: μ≤3.4 Ha: μ>3.4 E. H0: μ>3.3 Ha: μ≤3.3 F. H0: μ≠3.3 Ha: μ=3.3 (b) Identify the standardized test statistic. z=negative 0.93−0.93 (Round to two decimal places as needed.) (c) Find the P-value. 0.3520.352 (Round to three decimal places as needed.) (d) Decide whether to reject or fail to reject the null hypothesis.Today, the waves are crashing onto the beach every 5.6 seconds. The times from when a person arrives at the shoreline until a crashing wave is observed follows a Uniform distribution from 0 to 5.6 seconds. Round to 4 decimal places where possible. a. The mean of this distribution is b. The standard deviation is c. The probability that wave will crash onto the beach exactly 0.7 seconds after the person arrives is P(x = 0.7) = d. The probability that the wave will crash the beach between 1.6 and 5.1 seconds after the person arrives is P(1.6 3.72) = f. Suppose that the person has already been standing at the shoreline for 0.8 seconds without a wave crashing in. Find the probability that it will take between 1.4 and 3.3 seconds for the wave to crash onto the shoreline. g. 65% of the time a person will wait at least how long before the wave crashes in? seconds. h. Find the minimum for the upper quartile. seconds.
- Assume that the amounts of weight that male college students gain during their freshman year are normally distributed with a mean of μ = 1.1 kg and a standard deviation of o=4.9 kg. Complete parts (a) through (c) below. a. If 1 male college student is randomly selected, find the probability that he gains between 0 kg and 3 kg during freshman year. The probability is (Round to four decimal places as needed.) -CA journal published a study of the surface roughness of water pipes. An instrument measured the surface roughness (in micrometers) of 40 sampled sections of coated interior pipe. Let x¯x¯ denote the sample mean. Assume that the surface roughness distribution has a mean of μ=1.65μ=1.65 micrometers and a standard deviation of σ=0.8σ=0.8 micrometers. Find the probability that x¯x¯ exceeds 1.75 micrometers. Select one: a. 0.2148 b. 0.2683 c. 0.3979 d. 0.4356The undergraduate grade point averages (UGPA) of students taking an admissions test in a recent year can be approximated by a normal distribution, as shown in the figure. (a) What is the minimum UGPA that would still place a student in the top 10% of UGPAs? (b) Between what two values does the middle 50% of the UGPAs lie? 3.4042.8Grade point average μ=3.40 σ=0.17 x A normal curve labeled mu = 3.40 and sigma = 0.17 is over a horizontal x-axis labeled Grade point average from 2.8 to 4 in increments of 0.3 and is centered on 3.40. (a) The minimum UGPA that would still place a student in the top 10% of UGPAs is nothing. (Round to two decimal places as needed.) (b) The middle 50% of UGPAs lies between nothing on the low end and nothing on the high end. (Round to two decimal places as needed.)
- The round off errors when measuring the distance that a long jumper has jumped is uniformly distributed between 0 and 4.7 mm. Round values to 4 decimal places when possible. The mean of this distribution is The standard deviation is The probability that the round off error for a jumper's distance is exactly 1.1 is P(x = 1.1) = The probability that the round off error for the distance that a long jumper has jumped is between 0 and 4.7 mm is P(0.2 < x < 0.9) = The probability that the jump's round off error is greater than 1.04 is P(x > 1.04) = P(x > 2.8 | x > 0.6) = Find the 81st percentile. Find the maximum for the lower quartile.The undergraduate grade point averages (UGPA) of students taking an admissions test in a recent year can be approximated by a normal distribution, as shown in the figure. (a) What is the minimum UGPA that would still place a student in the top 15% of UGPAs? (b) Between what two values does the middle 50% of the UGPAs lie? 3.3042.6Grade point average μ=3.30 σ=0.16 x A normal curve labeled mu = 3.30 and sigma = 0.16 is over a horizontal x-axis labeled Grade point average from 2.6 to 4 in increments of 0.35 and is centered on 3.30. (a) The minimum UGPA that would still place a student in the top 15% of UGPAs is enter your response here. (Round to two decimal places as needed.)A uniform distribution has parameters a = 2.5 and β = 9.6. Find the probability for a randomly chosen value being within 1.3 standard deviation of the mean. Round to 4 decimal places.