The article "A Probabilistic Analysis of Dissolved Oxygen-Biochemical Oxygen Demand Relationship in Streams"(J. Water Resources Control Fed., 1969: 73-90) reported data on the rate of oxygenation in streams in a certainregion. The sample mean and standard deviation were computed as = 0.173 and s= 0.066, respectively.We would like to know if a normal distribution could be a reasonable model for the data. To do this, we havepartitioned the quantitative variable into 5 discrete regions and counted the number of streams observed tohave oxygenation rates in the given interval. First determine the probability of falling into each region if thenormal distribution is the appropriate model. Then use those probabilities with the expected counts to conducta Chi-squared goodness of fit test. Use a = 0.05.Frequency1220231513Rate (per day)Below .100.100 below.150.150 - below.200.200 - below .250.250 or more

Given that, The data on the rate of oxygenation in streams in a certain region. Sample mean (x) = 0.173 Standard deviation (s) = 0.066 Assume the Level of significance (α) = 0.05 The five discrete regions along with frequencies are given. The hypothesis for the test is, H0: The normal distribution is a reasonable model for the data.H1: The normal distribution is not a reasonable model for the data. The test statistic follows χ2-distribution with k-1 degrees of freedom. where k = number of discrete regions = 5 df=k-1=5-1=4 Test statistic formula: χ2=∑(O-E)^2E where O is the observed frequency. E is the expected frequency. The formula for Ei is, Ei = n*pi where n = total frequency = 83 First, we need to find the probability for each region. By using the normal distribution, p1=P(X<0.100) = P(X-meansd<0.100-0.1730.066)=P(Z<-1.11) By using excel, p2=P(0.100<X<0.150)=P(0.100-0.1730.066<x-meansd<0.150-0.1730.066)=P(-1.11<Z<-0.35)=P(Z<-0.35)-P(Z<-1.11) By using excel, Similarly, the remaining probabilities can be found below: Now the expected values are, E1=np1 =0.1335*83=11.08046E2=np2=0.22967*83=19.0626E3=np3=0.295928*83=24.562E4=np4=0.219902*83=18.25191E5=np5=0.121*83=10.04304 The test statistic is, Test statistic = 1.67 Now next find P-value for taking a decision. Decision: P-value (0.7962)>α(0.05) Fail to reject H0 at a 5% level of significance. It can be concluded that the normal distribution is a reasonable model for the data.