Question

When a truckload of apples arrives at a packing​ plant, a random sample of 175 is selected and examined for​ bruises, discoloration, and other defects. The whole truckload will be rejected if more than 6​% of the sample is unsatisfactory. Suppose that in fact 10% of the apples on the truck do not meet the desired standard. What is the probability that the shipment will be accepted​ anyway?

​P(accepted)=??

​(Round to three decimal places as​ needed.)

Answer 1

Sample size , n = 175

apples on the truck do not meet the desired standard = 10% = 0.1

P[ apple do not meet the desired standard ] = 0.1

The whole truckload will be rejected if more than 6​% of the sample is unsatisfactory.

In numbers, 175*0.06 = 10.5 apples or 10 apples do not meet the desired standard then the whole truckload will be rejected.

the truck load will be accepted if 10 or less apples are below the desired standard out of 175.

X be the number of defected apples

We need to find P[ X <= 10 ]

X ~ B(175,0.1)

where, p = P[ defected apples ] = 0.1

P[ X <= 10 ] = P[ X = 0 ] + P[ X = 1 ] + P[ X = 2 ] + P[ X = 3 ] + P[ X = 4 ] + P[ X = 5 ] + P[ X = 6 ] + P[ X = 7 ] + P[ X = 8 ] + P[ X = 9 ] + P[ X = 10 ]

P[ X = 0 ] = 175C0*0.1^0*(1 - 0.1 )^(175 - 0 ) = 175C0*0.1^0*0.9^175 = 0

P[ X = 1 ] = 175C1*0.1^1*(1 - 0.1 )^(175 - 1 ) = 175C1*0.1^1*0.9^174 = 0

P[ X = 2 ] = 175C2*0.1^2*(1 - 0.1 )^(175 - 2 ) = 175C2*0.1^2*0.9^173 = 0

P[ X = 3 ] = 175C3*0.1^3*(1 - 0.1 )^(175 - 3 ) = 175C3*0.1^3*0.9^172 = 0

P[ X = 4 ] = 175C4*0.1^4*(1 - 0.1 )^(175 - 4 ) = 175C4*0.1^4*0.9^171 = 0

P[ X = 5 ] = 175C5*0.1^5*(1 - 0.1 )^(175 - 5 ) = 175C5*0.1^5*0.9^170 = 0.0002

P[ X = 6 ] = 175C6*0.1^6*(1 - 0.1 )^(175 - 6 ) = 175C6*0.1^6*0.9^169 = 0.0007

P[ X = 7 ] = 175C7*0.1^7*(1 - 0.1 )^(175 - 7 ) = 175C7*0.1^7*0.9^168 = 0.0018

P[ X = 8 ] = 175C8*0.1^8*(1 - 0.1 )^(175 - 8 ) = 175C8*0.1^8*0.9^167 = 0.0043

P[ X = 9 ] = 175C9*0.1^9*(1 - 0.1 )^(175 - 9 ) = 175C9*0.1^9*0.9^166 = 0.0087

P[ X = 10 ] = 175C10*0.1^10*(1 - 0.1 )^(175 - 10 ) = 175C10*0.1^10*0.9^165 = 0.0161

P[ X <= 10 ] = P[ X = 0 ] + P[ X = 1 ] + P[ X = 2 ] + P[ X = 3 ] + P[ X = 4 ] + P[ X = 5 ] + P[ X = 6 ] + P[ X = 7 ] + P[ X = 8 ] + P[ X = 9 ] + P[ X = 10 ]

P[ X <= 10 ] = 0 + 0.0002 + 0.0007 + 0.0018 + 0.0032 + 0.0087 + 0.0161

P[ X <= 10 ] = 0.0318

P[ X <= 10 ] = 0.032 ( three decimal places )

P[ X <= 10 ] = 3.2%

P[ accepted ] = 0.032