Give the marginal probability of (e) * X = number of Bars of Signal Strength Y = response time (Nearest second) 1 2 3 P(Y) 4 0.15 0.1 0.05 3 0.02 0.1 0.05 2 0.02 0.03 0.2 -- 1 0.01 0.02 0.25 -- P(X) e -- -- O 0.2 O 0.25 O 0.55 O None of the choices
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- A model for normal human body temperature, X, when measured orally in °F, is that it is normally distributed, X ~ N (98.2,0.5184). (i) According to the model, what proportion of people have a normal body temperature of 99 °F or more? (ii) Find the normal body temperature such that, according to the model, only 10% of people have a lower normal body temperature. (iii) Let W denote normal human body temperature, when measured orally in °C. Given that W = (X - 32) and X ~ N(98.2,0.5184), what is the distribution of W? (iv) According to the model that you just derived for W, what proportion of people have a normal body temperature of between 36 °C and 36.8 °C?H0:μ=12.8H0:μ=12.8H1:μ<12.8H1:μ<12.8Your sample consists of 45 subjects, with a mean of 12.3 and standard deviation of 1.27.Calculate the test statistic, rounded to 2 decimal places. You are conducting a study to see if the probability of a true negative on a test for a certain cancer is significantly less than 0.28. Your sample data produce the test statistic z=−2.672z Find the p-value accurate to 4 decimal places.A normal distribution has a mean of u = 50 and 0 = 12 if one score is randomly selected for distribution what is the probability of randomly selectingt a score that is X < 56
- The Table below shows the average score of the150 students sampled from the University of the East Europe in the academic year 2020/2021. Score (out of 100) Number of students 40-49 19 50-59 32 60-69 49 70-79 29 80-89 21 Given that Class Frequency 40 - 49 19 50 - 59 32 60 - 69 49 70 - 79 29 80 - 89 21 Class(1) Frequency (f)(2) Mid value (x)(3) f⋅x(4)=(2)×(3) cf(6) 40-49 19 44.5 845.5 19 50-59 32 54.5 1744 51 60-69 49 64.5 3160.5 100 70-79 29 74.5 2160.5 129 80-89 21 84.5 1774.5 150 n=150 ∑f⋅x=9685 Use your results above to comment on the shape of the distribution. Give a reason for you answer.llMetro by T-Mobile X Group Work-Worksheet#1 on Ch... 1 of 2 Group Work-Worksheet#1 on Chapter 3 0.14% → 1. The Empirical Rule Based on Data Set 1" Body Data" in appendix B, blood platelet counts of women have a normal/bell-shaped distribution with a mean of 255.1 and a standard deviation of 65.4. (All units are 1000 cells/μL.) Using the empirical rule: -30 2.14% 1:40 PM 13.59% 99.73% 95.44% 68.26% 34.13% 34.13% 13.59% 77% 2.14% -20 -10 Mean lo 20 За a. 68.26% of women with platelet counts are within two standard deviations of the mean? The values are from ( b. 95.44% of women with platelet counts are within two standard deviations of the mean? The values are from ( 000 c. 99.73% of women with platelet counts are within two standard deviations of the mean? The values are from ( 0.14%I need help with this question, A and C are wrong and I don't know D
- A normal distribution has a mean of μμ= 54 and a standard deviation of σσ=6 What is the probability of randomly selecting a score GREATER than 51? A normal distribution has a mean of μμ= 54 and a standard deviation of σσ=6 What is the probability of selecting a sample of n = 4 scores with a mean more than M = 51? A normal distribution has a mean of μμ= 54 and a standard deviation of σσ=6 What is the probability of selecting a sample of n = 36 scores with a mean more than M = 51?7/8/22, 3:04 PM Area z TABLE V z 78877 99737 77238 23927 328797 97979 777738=283 09898 30003 99999 33233 38539 22 -3.4 -2.5 -0.5 0.0010 0.0013 .00 .01 .02 .06 .07 .08 .09 0.0007 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0002 0.0005 0.0005 0.0005 0.0004 0.0004 0.0004 0.0004 0.0004 0.0004 0.0003 0.0007 0.0006 0.0006 0.0006 0.0006 0.0006 0.0005 0.0005 0.0005 0.0009 0.0009 0.0009 0.0008 0.0008 0.0008 0.0008 0.0007 0.0007 0.0013 0.0013 0.0012 0.0012 0.0011 0.0011 0.0011 0.0010 0.0010 0.0019 0.0018 0.0018 0.0017 0.0016 0.0016 0.0015 0.0015 0.0014 0.0014 0.0026 0.0025 0.0024 0.0023 0.0023 0.0022 0.0021 0.0021 0.0020 0.0019 0.0035 0.0034 0.0033 0.0032 0.0031 0.0030 0.0029 0.0028 0.0027 0.0026 0.0047 0.0045 0.0044 0.0043 0.0041 0.0040 0.0039 0.0038 0.0037 0.0036 0.0062 0.0060 0.0059 0.0057 0.0055 0.0054 0.0052 0.0051 0.0049 0.0048 0.0082 0.0080 0.0078 0.0075 0.0073 0.0071 0.0069 0.0068 0.0066 0.0064 0.0107 0.0104 0.0102 0.0099 0.0096 0.0094 0.0091 0.0089 0.0087 0.0084…Question * Given X be the life time of a bulb having a probability density function f(x) =-e for x>0 and 0 elsewhere. The mean life time of the bulbs is equal to: %3D O 1/2 O None of these O 1/4 Question* Let X and Y be two independent continuous random variables with marginal distribution
- 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 42Explain into detial and show all steps for me to understand how to do it by myself please. Thank you.A population of values has a normal distribution with μ=184.5μ=184.5 and σ=15.5σ=15.5. You intend to draw a random sample of size n=68n=68.Find P40, which is the mean separating the bottom 40% means from the top 60% means.P40 (for sample means) = Enter your answers as numbers accurate to 1 decimal place. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.