The average household of four has about 2,000 lbs of laundry in a year. Suppose this...

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.

Solutions

Expert Solution

(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

orchestra answered 2 years ago

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