**6.30 Fuel Efficiency**Computers in some vehicles calculate various quantities related to performance. One of these is the fuel efficiency, or gas mileage, usually expressed as miles per gallon (mpg). For one vehicle equipped in this way, the miles per gallon were recorded each time the gas tank was filled, and the computer was then reset. Here are the mpg values for a random sample of 20 of these records: 41.5, 50.7, 36.6, 37.3, 34.2, 45.0, 48.0, 43.2, 47.7, 42.2, 43.2, 44.6, 48.4, 46.4, 46.8, 39.2, 37.3, 43.5, 44.3, 43.3Suppose that the standard deviation is known to be $\sigma = 3.5$ mpg.(a) **What is $\sigma_x$, the standard deviation of $\bar{x}$?**(b) **Examine the data for skewness and other signs of non-Normality. Show your plots and numerical summaries. Do you think it is reasonable to construct a confidence interval based on the Normal distribution? Explain your answer.**(c) **Give a 95% confidence interval for $\mu$, the mean miles per gallon for this vehicle.**

The given sample observations are: 41.5 50.7 36.6 37.3 34.2 45 48 43.2 47.7 42.2 43.2 44.6 48.4 46.4 46.8 39.2 37.3 43.5 44.3 43.3 Part (a) The standard deviation of x¯ is calculated as: σx¯=σn=3.520=0.175. Part (b) Of the obtained 20 sample observations, it is calculated that: mean, x¯=41.5+50.7+...+43.320=43.17 Considering the observations in ascending order, it is obtained as: 34.2 36.6 37.3 37.3 39.2 41.5 42.2 43.2 43.2 43.3 43.5 44.3 44.6 45 46.4 46.8 47.7 48 48.4 50.7 It is calculated that (considering n is even): Q1 (first quartile)=n+14th observation=20+14th observation=5.25th value=5th observation+0.256th observation-5th observation=39.2+0.2541.5-39.2=39.775 Q2 (second quartile)=n2th observation value+n2+1th observation value2=202th observation value+202+1th observation value2=10th value+11th value2=43.3+43.52=43.4 Q3 (third quartile)=3n+14th observation=320+14th observation=15.75th value=15th observation+0.7516th observation-15th observation=46.4+0.7546.8-46.4=46.7 The Bowley's Coefficient of Skewness is: Sk=Q3+Q1-2Q2Q3-Q1=46.7+39.775+2×43.446.7-39.775≈-0.047 Considering the skewness coefficient is very close to 0 (the value that defines the perfect symmetricity of a distribution), the sample is considered to be approximately symmetrical. Moreover, since the x¯=43.17≈Q243.4, the distribution is approximately symmetric about the mean (and median), hence the sample distribution may be considered to be approximately normal. The box plot is given as: Since the sample distribution is inferred to be normally distributed, the CI for the true mean miles per gallon may be derived using the provided sample observation. Part (c) Considering the standard deviation of the distribution of miles per gallon is specified at the population level, i.e., σ=3.5, the z-critical value is to be considered for defining the true average miles per gallon value (i.e., μ). For the given confidence level of 95% (i.e., α=1-95%=0.05), the z-critical value is obtained using the excel function "NORM.S.INVprobability" as follows (where probability=1-α2=1-0.052=0.975): So the z-critical value for 95% confidence is 1.96. And the CI is obtained as: x¯±zσx¯=43.17±1.96×0.175 from part-a=43.17±0.343i.e., 43.17-0.343, 43.17+0.343=42.827, 43.513 With 95% confidence,, the value of μ is claimed to lie between 42.827 and 43.513 mpg.