Respuesta :
Answer:
c) P(270≤x≤280)=0.572
d) P(x=280)=0.091
Step-by-step explanation:
The population of bearings have a proportion p=0.90 of satisfactory thickness.
The shipments will be treated as random samples, of size n=500, taken out of the population of bearings.
As the sample size is big, we will model the amount of satisfactory bearings per shipment as a normally distributed variable (if the sample was small, a binomial distirbution would be more precise and appropiate).
The mean of this distribution will be:
[tex]\mu_s=np=500*0.90=450[/tex]
The standard deviation will be:
[tex]\sigma_s=\sqrt{np(1-p)}=\sqrt{500*0.90*0.10}=\sqrt{45}=6.7[/tex]
We can calculate the probability that a shipment is acceptable (at least 440 bearings meet the specification) calculating the z-score for X=440 and then the probability of this z-score:
[tex]z=(x-\mu_s)/\sigma_s=(440-450)/6.7=-10/6.7=-1.49\\\\P(z>-1.49)=0.932[/tex]
Now, we have to create a new sampling distribution for the shipments. The size is n=300 and p=0.932.
The mean of this sampling distribution is:
[tex]\mu=np=300*0.932=279.6[/tex]
The standard deviation will be:
[tex]\sigma=\sqrt{np(1-p)}=\sqrt{300*0.932*0.068}=\sqrt{19}=4.36[/tex]
c) The probability that between 270 and 280 out of 300 shipments are acceptable can be calculated with the z-score and using the continuity factor, as this is modeled as a continuos variable:
[tex]P(270\leq x\leq280)=P(269.5<x<280.5)\\\\\\z_1=(x_1-\mu)/\sigma=(269.5-279.6)/4.36=-10/4.36=-2.29\\\\z_2=(x_2-\mu)/\sigma=(280.5-279.6)/4.36=0.9/4.36=0.21\\\\\\P(270\leq x\leq280)=P(-2.29<z<0.21)=P(z<0.21)-P(z<-2.29)\\\\P(270\leq x\leq280)=0.583-0.011=0.572[/tex]
d) The probability that 280 out of 300 shipments are acceptable can be calculated using again the continuity factor correction:
[tex]P(X=280)=P(279.5<X<280.5)\\\\\\z_1=(x_1-\mu)/\sigma=(279.5-279.6)/4.36=-0.1/4.36=-0.02\\\\z_2=(x_2-\mu)/\sigma=(280.5-279.6)/4.36=0.9/4.36=0.21\\\\\\P(X=280)=P(-0.02<z<0.21)=P(z<0.21)-P(z<-0.02)\\\\P(X=280)=0.583-0.492=0.091[/tex]
