Question

Instructions: For full credit, you must clearly illustrate each problem on the graph provided, with labels and using terminology similar to that used in class: F(T), F(Z), a, P(X>value), P(X=value), etc. Clearly show and label the mean, std dev, and the significant values and areas needed to support your answer. Show any computations necessary to support your answer in the space provided. Print clearly. Provide answers to 3-decimal place accuracy (1-decimal for percentages). Note: you must show interpolation for problem 4; on other problems, round to nearest value on distribution table. Hint: answers to 1 & 2 and 3 & 4 should be similar but not exactly equal. The lifespan of a new LED bulb is being studied. Testing has determined the lifespan is approximately normally distributed with an average of 5000 hours and a standard deviation of 75 hours.

Answer 1

-0. 85 Sooo 50777 0.85 solution Here mean (4) = 5000, S.D(+) = 25 D The z- valce at 85th percentile is 1.036 - P( Z21.036)=0.85 them 85th percentile is U+10360 = 5000 +1.036 (15) = 50777 2 here i-e. p(x2 5077.7) = 0.85 2 here n=9, df = n-1=8 PA21:108) = 0.85 The 85th porcentile es U +1.1086 = 5000 + 1-108(75) = 5083-1 ie. P(x 250831=0.85 3 for x= uqoo, Z=3y = 1900 – 5000 = -33 Plabove 4900) =pl&> 4900) 5000 5083 4900 5000 =P(27-1.33) =1-P124-133) =1-0.0918 = 0.9082 = 90.82%