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An article the ASCE Joumnal Energy Engincering (Vol 125. 1999 59-75 decibc study the thermal mertid prupETTICS 4uloclared hcrhteL cuncTete ICO builliug material F...

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An article the ASCE Joumnal Energy Engincering (Vol 125. 1999 59-75 decibc study the thermal mertid prupETTICS 4uloclared hcrhteL cuncTete ICO builliug material Five" samples of the material WVCc tested StruCLuIC and the average interior temperatun C) reported follors: 23.0. 22.22,22.04. 22.62. aud 22.59.The parameter of interect [JCHIM mtenur tempCTature Test bypothezis 22.5 FeTsus H : [ F 22.5 . using 0.05. Use the -nlue appruuch_(b) Find 95% confidence interwal on the IeaIL MItETIUI tc

An article the ASCE Joumnal Energy Engincering (Vol 125. 1999 59-75 decibc study the thermal mertid prupETTICS 4uloclared hcrhteL cuncTete ICO builliug material Five" samples of the material WVCc tested StruCLuIC and the average interior temperatun C) reported follors: 23.0. 22.22,22.04. 22.62. aud 22.59. The parameter of interect [JCHIM mtenur tempCTature Test bypothezis 22.5 FeTsus H : [ F 22.5 . using 0.05. Use the -nlue appruuch_ (b) Find 95% confidence interwal on the IeaIL MItETIUI tcmpeTALIC



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Two methods were used to study the latent heat of ice fusion. Both method $A$ (an electrical method) and method $\mathrm{B}$ (a method of mixtures) were conducted with the specimens cooled to $-0.72^{\circ} \mathrm{C}$. The data in the following table represent the change in total heat from $0.72^{\circ} \mathrm{C}$ to water at $0^{\circ} \mathrm{C}$ in calories per gram of mass. Assuming normality, construct a $95 \%$ confidence interval for the difference between the means.

Were given a log. Normal distribution with mu equals 4.5 and sigma equals 0.8 for part A were asked to find the mean and standard deviation of the distribution. The meat, the formula for the mean is given as follows. So if we plug in our parameter values, this comes out to 123.97 and the formula for the variance is given here. And this comes out to 13,776 0.53 which gives us a standard deviation of 117.37 And so we have our mean and standard deviation for the distribution for B were asked for the probability that X is at most 100. This is equal to the probability that the natural algorithm of X is less than or equal to the natural algorithm of 100. And then if we standardize, this is equal to the cumulative probability of the standard normal distribution at 0.131 and that comes out to a probability of 0.552 Now for part C. We're asked for the probability that X is at least 200 and greater than 200 now, because X is a continuous distribution, the probability that X is at least 200 is equal to the probability that X exceeds 200. And this is equal to one, minus the probability that is, at most 200. And then if we standardize, yeah, we get the following. This is taking the log rhythm and standardizing in one step. This comes out to one minus 0.841 or a 0.159 So the probability that X is at least 200 is 0.159

16. It's standard deviation, uh is equal to the square. Root off the sum of the square deviation divided by end minus one, which is equal to open toe 51 toe. Remind the critical values from Table six the degree the freedom is equal to 14, which is in minus 1 50 minus one. So the chi square off one minus open toe find is equal to 5.6 to 9, and the chi square, off open or five, is equal to 26 for 119 So the boundaries for the standard deviation is equal to square. Root off N minus one over chi square off Alfa over to times s Geo 15 minus 1/26 point point 19 times opened over 51 which approximately equal toe operate or 37 And the square root of n minus one over X squared off one minus off over two times. Asked which is 15 minus 1/5 150.6 to 9 times opened, or 51 which approximately offering or eight. So the boundaries for the variants are the square off these venues, which is opened or 37 square and open 0, 80 square, which is a kind 014 square, and or find 06 cool

In this exercise, we have the time to failure described by a Waibel distribution with shape parameter 2.5 and scale parameter 200 for part A were asked for the probability that the lifetime is at most 250. We're looking for the probability of tea is at most 250. I recall that the CDF for a viable distribution has given as follows, which means that we can write this probability as follows and this comes out to 0.826 in this question were also asked for. The probability at the lifetime is strictly less than 250 since a Waibel random variable is a continuous random variable. This is also equal to the probability that the lifetime is at most 2 50 which we just for solved for 0.826 And the third part of part A is the probability that the lifetime exceeds 300. So, using the formula for the CDF for Waibel, we end up with 0.64 The probability that the lifetime exceeds 300 hours is about 0.64 now moving on to part B were asked for the probability that the lifetime is between 102 150 hours. This is equal to the cumulative probability at 250 minus the cumulative probability at 100. Now, we've already solved for the first term here, So this is equal to 0.826 This is equal to 0.664 and then for part C, we're asked for the value for which 50% of the lifetimes would exceed. This is we're looking for the 50% Kwan tile, which is equal to the inverse CDF of 0.5. So the CDF of the 50% out the 50th percentile is equal. Thio 0.5. That is the cumulative problem. The cumulative probability of the 50th percentile is equal to 0.5. So we're looking to solve for T in this equation. So we have t to explain it. 200 30 divided by 200 to the explain it 2.5 is equal to minus the natural logarithms of 0.5 and this gives us a value of T equals 172 0.73 hours, so 50% of the time, the lifetime will be more than 172.73 and 50% of the time it will be less.


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