I'm having problems mostly with part (b) and (c). Please add every single step and side note you can - including formulas used. I greatly appreciate the help!

A company purchases large shipments of lemons and uses this acceptance smaplong plan: randomly select and test 200 lemons, than accept the whole batch if there is less than two found to be rotten bitter; that is, at most one lemon is bad. If a particular shipment actually has a 3% rate of defects:

a. Using the binomial probability distribution, find the probability that this whole shipment is accepted?

P(Accepted) = P(0) + P(1) = ( 200C0 * (0.03)^0 * (0.97)^200 ) + ( 200C1 * (0.03)^1 * (0.97)^199 ) = 0.0162

b. Show that a normal approximation to this binomial distribution is appropriate.

np = 6 => 5, n(1-p) = 194 => 5

c. Using normal approximation (either with or without continuity correction), find the probability that this whole shipment is accepted?

w/o correction: P(Accepted) = P(X <= 1 | μ = 6, σ^2 = 5.82) = P(Z <= -2.07) = 0.0191

w/ correction: P(Accepted) = P(X <= 1.5 | μ = 6, σ^2 = 5.82) = P(Z <= -1.87) = 0.0311

note: <= stands for 'less than or equal to'

A company purchases large shipments of lemons and uses this acceptance smaplong plan: randomly select and test 200 lemons, than accept the whole batch if there is less than two found to be rotten bitter; that is, at most one lemon is bad. If a particular shipment actually has a 3% rate of defects:

a. Using the binomial probability distribution, find the probability that this whole shipment is accepted?

P(Accepted) = P(0) + P(1) = ( 200C0 * (0.03)^0 * (0.97)^200 ) + ( 200C1 * (0.03)^1 * (0.97)^199 ) = 0.0162

b. Show that a normal approximation to this binomial distribution is appropriate.

np = 6 => 5, n(1-p) = 194 => 5

c. Using normal approximation (either with or without continuity correction), find the probability that this whole shipment is accepted?

w/o correction: P(Accepted) = P(X <= 1 | μ = 6, σ^2 = 5.82) = P(Z <= -2.07) = 0.0191

w/ correction: P(Accepted) = P(X <= 1.5 | μ = 6, σ^2 = 5.82) = P(Z <= -1.87) = 0.0311

note: <= stands for 'less than or equal to'

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