Reference information for the question:For a certain type of salamander, the mass (X) of a randomly chosen adult is normally distributed.95% of adults in this species have a mass between 6.7 grams and 8.1 grams. Salamanders Again!Fo.•Ction, refer to the Week 6 CollaborDescribe the distribution of samples of size n =100 from this population.Sketch this sampling distribution, and also sketch the distribution of salamander masses. Compare andcontrast the two distributions.What is the probability of getting a sample of size n100 with mean < 7.2? Explain your process.P• What is the probability of getting an individual salamander with a mass of X < 7.2 grams? Explain yourO●process.Explain the difference between these two computations, and illustrate on the graphs of the twodistributions.Without actually calculating it, is the probability of getting a sample of size n200 with meanX < 7.2 higher or lower than the probability of getting the same mean for a sample of size n = 100?Explain your reasoning.CJUsignment, the scenario with the salamanders:31acerOllOVAUSNov 27

The mass X of a randomly chosen adult is normally distributed. A 95% confidence interval for the mean mean mass of that species of salamander is (6.7, 8.1).The mean and standard deviation of X is calculated as follows: μ=8.1+6.72=7.4 From the Empirical rule, we know that 95% of the observed data lies within 2 standard deviations from the mean. So, we can write: μ-2σ=6.77.4-2σ=6.7σ=7.4-6.72=0.35 Hence, the population mean is 7.4 grams and population standard deviation is 0.35 grams.Sampling distribution for a sample of size 100: When the population is normally distributed, the sampling distribution of sample mean follows normal distribution with sample mean x¯ =μ and sample standard deviation s=σn Therefore, x¯=7.4 grams and s=0.35100=0.035To sketch the sampling distribution, use the following commands in R programming: # Mean and s.d mu <− 7.4 s.d <− 0.035 #Sequence of random numbers x <− seq(−4,4,length=1000)*s.d + mu #PDF of random numbers y <− dnorm(x, mu, s.d) #plot plot(x,y, type = "l", main = "Sampling Distribution") The sketch is as follows: To sketch the distribution of salamander masses, consider the following: # Mean and s.d mu <− 7.4 s.d <− 0.35 #Sequence of random numbers x <− seq(−4,4,length=1000)*s.d + mu #PDF of random numbers y <− dnorm(x, mu, s.d) #plot plot(x,y, type = "l", main = "Population Distribution") From the sketch of population and sampling distribution, the shape of both the distribution is bell shaped. The sampling distribution has larger tails than the population distribution, implying that the sampling distribution has more area in the tails as compared to the population distribution. The sampling distribution is more platykurtic than the population curve.We know that X¯~ N7.4, s=0.035 PX¯<7.2=PX¯-μσn<7.2-7.40.035=PZ<-5.71≅0 Hence, the probability of getting a sample of size 100 with X¯<7.2 is almost negligible. The normal probability can be calculated using the Excel function =NORM.S.DIST(-5.71,TRUE)We know that for individual salamander, X~N7.4, σ=0.35 PX<7.2=PX-μσ<7.2-7.40.35=PZ<-.571=0.284 Hence, the probability of getting a salamander with X<7.2 is 0.284 The normal probability can be calculated using the Excel function =NORM.S.DIST(-0.571,TRUE)General Rule: As the sample size increase, the observed values move closer to the true population mean. When we compute X¯<7.2, we know that for a sample mean the values will lie very close to the mean of the distribution. So, there will be very less area to the left of 7.2 When we compute X<7.2, we know that for a single value the standard deviation will be large. So, there will be more area to the left of 7.2As we know that as sample size increases, the standard error of sample mean decreases and the observed values come close to the true population parameter. Therefore, the probability of X¯<7.2 with n=200 will be less than when n=100 because the sample size of 200 is larger than the sample size of 100.