In: Statistics and Probability
The average household of four has about 2,000 lbs of laundry in a year. Suppose this data is normally distributed with a standard deviation of 187.5. Find the probability that a random sample of 49 families will have less than 1980 lbs. of laundry per year. Round the final answers to at least four decimal places and intermediate z-value calculations to two decimal places.
A.0.3409
B.0.2266
C.0.6591
D.0.7734
2. Find critical z-value for 92% confidence interval. Answer should have a positive value with 2 decimal places.
3. Find critical z-value for 84% confidence interval. Answer should have a positive value with 2 decimal places.
(1)
= / n = 187.5 / 49
P( < 1980) = P(( - ) / < (1980 - 2000) / 187.5 / 49 )
= P(z < -0.75)
= 0.2266
(2)
At 92% confidence level the z is ,
= 1 - 92% = 1 - 0.92 = 0.08
/ 2 = 0.08 / 2 = 0.04
Z/2 = Z 0.04 = 1.75
(3)
At 82% confidence level the z is ,
Z/2 = Z 0.09 = 1.34