Question

Question The number of insurance claims that a restaurant submits each year has a normal distribution with a population stand deviation of 0.9 and an unknown population mean. If a random sample of 54 restaurants is taken and results in a sam mean of 3.3 claims, find a 95% confidence interval for the population mean. Use the tool below to calculate your confidence interval. Choose the correct confidence level, sample mean, population standard deviation, and sample size. Then view the confidence interval indicated on the x-axis to find your answer. Pleas note, you'll need to use the sliders above the graph to change the x-axis markers. Round your answers to 2 decimal place

Answer 1

If x is rondom v mean at Confidence interval for population 95% of normal dietribution ME X + 1.165] ① given I = 3.3 J = 0.9 n = 54 Coryidence interval for population meam dy ł - 1.965 < au < ñ + 1,96€ < a < 33 + 1.968 3.3 - 1.968 0.9 54 3.3 – 0.24 0.g. 154 <ul 3.3+ 0.24 3.06 < 1 3.54) ut [3.06, 3.54] m=37 a= 3.5 T = 0.6 confidence interval for pot" meon at 997_o nomaldist MET X + 2580 [- 2.58 7 <us x + 215857 3.5 - 2.50xan Zee 3.5 +2.58 x 047 T 3.5 0.25 < uk 3.5 +0.25 3.25 < u 3.75