**Analyzing Helium Porosity of Coal Samples**Assume that the helium porosity (in percentage) of coal samples taken from any particular seam is normally distributed with a true standard deviation of **0.70**.**(a)** Compute a 95% Confidence Interval (CI) for the true average porosity of a certain seam if the average porosity for **17** specimens from the seam was **4.85**. (Round your answers to two decimal places.)\[ \text{( } \_\_\_\_ , \_\_\_\_ \text{ )} \]**(b)** Compute a 98% Confidence Interval (CI) for the true average porosity of another seam based on **17** specimens with a sample average porosity of **4.56**. (Round your answers to two decimal places.)\[ \text{( } \_\_\_\_ , \_\_\_\_ \text{ )} \]**(c)** How large a sample size is necessary if the width of the 95% interval is to be **0.34**? (Round your answer up to the nearest whole number.)\[ \_\_\_\_ \text{ specimens} \]**(d)** What sample size is necessary to estimate true average porosity to within **0.23** with 99% confidence? (Round your answer up to the nearest whole number.)\[ \_\_\_\_ \text{ specimens} \]

It is given that the random variable is normally distributed with true standard deviation σ = 0.70. Note: According to Bartleby expert guidelines, we can answer only first question with their three subparts and rest can be reposted.Solution(a): The critical value for α = 0.05 for two-tailed test is zc = 1.96 The 95% confidence interval is, C.I = ( x̄ ± zc × σ/√n ) = ( 4.85 ± 1.96 × 0.70/√17 ) = ( 4.52, 5.18) Note: The Excel function is:=NORM.S.INV(1-0.05/2) The Excel output is:Solution(b): The critical value for α = 0.02 for two-tailed test is zc = 2.326 The 98% confidence interval is, C.I = ( x̄ ± zc × σ/√n ) = ( 4.56 ± 2.326 × 0.70/√17 ) = ( 4.17, 4.96) Note: The Excel function is:=NORM.S.INV(1-0.02/2) The Excel output is: