Question #88279
hospital records show that of the patients suffering from a certain disease only 30% recover
a) what is the probability that of 8 randomly selected patients of this disease exactly 5 will recover
b) explain why this scenario depicts a bionomial process
c) the weight of wheat bags produced by pemble ltd follows a normal distribution with a mean of 80kg and a standard deviation of 8kg find the probability that a randomly selected bag will weigh
a) more than 120 bags
b) less than 90 bags
c) if a whole sale trader buys 50bags from the factory, find how many will be between 66kg and 82kg
1
Expert's answer
2019-04-22T12:34:25-0400

1.

a)

P(x=k)=Cnkpk(1p)(nk)P(x=k)=C_n^kp^k(1-p)^(n-k)


p(x=5)=C850.35(10.3)3=8!/(3!5!)0.00083=0.0465p(x=5)=C_8^5*0.3^5*(1-0.3)^3=8!/(3!5!)*0.00083=0.0465

b) This scenario depicts a binomial process because:
we have the sequence of n independent experiments (selected patients), 
each experiment has two outcomes (recover, not recover), and know probabilities of these outcomes(p,(1-p))


2.

a)

z=(xμ)/σz=(x-\mu)/\sigma

z=(12080)/8=5z=(120-80)/8=5

P(x>120kg)=P(z>5)=1P(z<5)=10.9999=0.0001P(x>120kg)=P(z>5)=1-P(z<5)=1-0.9999=0.0001



b)


z=(9080)/8=1.25z=(90-80)/8=1.25


P(x<90kg)=P(z<1.25)=0.8944P(x<90kg)=P(z<1.25)=0.8944

c)


z=(xμ)/(σ/n)z=(x-\mu)/(\sigma/\sqrt{n})

z1=(6680)/(8/50)=0.07z_1=(66-80)/(8/\sqrt{50})=-0.07

z2=(8280)/(8/50)=0.04z_2=(82-80)/(8/\sqrt{50})=0.04

P(66<x<82)=P(z<0.04)P(z<0.07)=0.51600.4721=0.0439P(66<x<82)=P(z<0.04)-P(z<-0.07)=0.5160-0.4721=0.0439


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