An article suggests the uniform distribution on the interval (7.5, 20) as a model for depth (cm) of the bioturbation layer in sediment in a certain region.(a) What are the mean and variance of depth? (Round your variance to two decimal places.)mean13.7513.02variance(b) What is the cdf of depth?0F(x) =x-7.512.51x < 7.57.5 < x < 2020 ≤ x(c) What is the probability that observed depth is at most 10? (Round your answer to four decimal places.)0.20What is the probability that observed depth is between 10 and 15? (Round your answer to four decimal places.)0.40(d) What is the probability that the observed depth is within 1 standard deviation of the mean value? (Round your answer to four decimal places.)What is the probability that the observed depth is within 2 standard deviations of the mean value?

Given that X ~ Uniform ( 7.5 , 20 ) Note : According to bartleby experts question answers…

Answered: (a) What are the mean and variance of…

A First Course in Probability (10th Edition)

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Chapter1: Combinatorial Analysis

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rse ations Suppose a study was done to determine whether the average amount of sleep that students get the night before an exam is less than 6 hours. An SRS of 100 students from a university was taken, and a mean of 5.5 hours was computed from the sample. The figure below provides the results of the analysis. Normal calculator Density 04 0.3 02 0.1 0 -3 -2 Mean: 0 Std. Dev. 1 Prob(X<= -2.5 )=0.0062096654 Snapshot Close Compute What is your conclusion based on the results of this study? (Assume a significance level of 0.05.) O On average, students typically get less than 6 hours of sleep the night before an exam. O On average, students typically get 6 hours of sleep the night before an exam. O On average, students typically get more than 6 hours of sleep the night before an exam. O On average, students typically get 5.5 hours of sleep the night before an exam.
Hocument/d/1ho6gSyZTqOTIUok211PsszM1fL7Jz4K2HPR4WWL4x1g/edit sert Format Tools Extensions Help 2 100% D 1 ▼ Normal text 2 1 W L #3 E 2 $ Arial A quality control expert at Glotech computers wants to test their new monitors. The production manager claims they have a mean life of 83 months with a variance of 100. If the claim is true, what is the probability that the mean monitor life would be greater than 81.3 months in a sample of 89 monitors? Round your answer to four decimal places. R % 5 T ^ 10.5 + 6 Y & B 7 * CO I U 5 8 I A ( 9 ) 0 6 P
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H0 : =150.00 kPa H1 :>150.00 kPa b) α= 0.01; sample variance : 1300 (freq. table ) sample deviation : 36.0555 (freq. table ) Calculation of Data Mean of the results= = 6700/50 =134 Mid Range of the results= (217.32+46.32)/2= 131.82 Range = (217.32-46.32) = 171 Modal Class= Class 121-145 with the highest frequency of 15 Median of the sorted set of data = 132.25 Mode = there are not clear modes In the samples. Sample variance S o= 1300 Sample deviation: S =36.55
What is the value of the variance of weight gain for the off-campus students? (Round your answer to two decimal places.)
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Data of variable-A is collected as shown to the right. Is the mean of Variable-A less than 80? State null hypothesis and choose level of significance, a = 0.05. (Use Excel ttest function, provide clear logic and reasoning.) Variable-A Variable-B 78.8 81.9 76.9 78.0 67.8 69.7 68.9 74.7 70.5 70.0 80.0 68.2 63.0 88.7 78.7 86.7 85.4 69.5 87.0 75.1 82.4 71.8 78.2 58.1 76.9 74.8 60.2 93.2 82.6 84.6 67.2 66.5 71.7 77.2 68.9 74.0 69.3 77.7 89.1 78.6 76.3 68.3 74.0 89.0 69.4 82.1 76.3 78.0 66.2 80.3 73.9 82.4 79.1 83.1 80.4 66.4 66.2 67.6 80.9 75.3
This output represents the hypothesis test that the average age at first pregnancy of women diagnosed with breast cancer is different depending on whether or not metastases are present (yes or no) at the time of breast cancer diagnosis.
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