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:
![\mu_s=np=500*0.90=450](https://tex.z-dn.net/?f=%5Cmu_s%3Dnp%3D500%2A0.90%3D450)
The standard deviation will be:
![\sigma_s=\sqrt{np(1-p)}=\sqrt{500*0.90*0.10}=\sqrt{45}=6.7](https://tex.z-dn.net/?f=%5Csigma_s%3D%5Csqrt%7Bnp%281-p%29%7D%3D%5Csqrt%7B500%2A0.90%2A0.10%7D%3D%5Csqrt%7B45%7D%3D6.7)
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:
![z=(x-\mu_s)/\sigma_s=(440-450)/6.7=-10/6.7=-1.49\\\\P(z>-1.49)=0.932](https://tex.z-dn.net/?f=z%3D%28x-%5Cmu_s%29%2F%5Csigma_s%3D%28440-450%29%2F6.7%3D-10%2F6.7%3D-1.49%5C%5C%5C%5CP%28z%3E-1.49%29%3D0.932)
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:
![\mu=np=300*0.932=279.6](https://tex.z-dn.net/?f=%5Cmu%3Dnp%3D300%2A0.932%3D279.6)
The standard deviation will be:
![\sigma=\sqrt{np(1-p)}=\sqrt{300*0.932*0.068}=\sqrt{19}=4.36](https://tex.z-dn.net/?f=%5Csigma%3D%5Csqrt%7Bnp%281-p%29%7D%3D%5Csqrt%7B300%2A0.932%2A0.068%7D%3D%5Csqrt%7B19%7D%3D4.36)
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:
![P(270\leq x\leq280)=P(269.5](https://tex.z-dn.net/?f=P%28270%5Cleq%20x%5Cleq280%29%3DP%28269.5%3Cx%3C280.5%29%5C%5C%5C%5C%5C%5Cz_1%3D%28x_1-%5Cmu%29%2F%5Csigma%3D%28269.5-279.6%29%2F4.36%3D-10%2F4.36%3D-2.29%5C%5C%5C%5Cz_2%3D%28x_2-%5Cmu%29%2F%5Csigma%3D%28280.5-279.6%29%2F4.36%3D0.9%2F4.36%3D0.21%5C%5C%5C%5C%5C%5CP%28270%5Cleq%20x%5Cleq280%29%3DP%28-2.29%3Cz%3C0.21%29%3DP%28z%3C0.21%29-P%28z%3C-2.29%29%5C%5C%5C%5CP%28270%5Cleq%20x%5Cleq280%29%3D0.583-0.011%3D0.572)
d) The probability that 280 out of 300 shipments are acceptable can be calculated using again the continuity factor correction:
![P(X=280)=P(279.5](https://tex.z-dn.net/?f=P%28X%3D280%29%3DP%28279.5%3CX%3C280.5%29%5C%5C%5C%5C%5C%5Cz_1%3D%28x_1-%5Cmu%29%2F%5Csigma%3D%28279.5-279.6%29%2F4.36%3D-0.1%2F4.36%3D-0.02%5C%5C%5C%5Cz_2%3D%28x_2-%5Cmu%29%2F%5Csigma%3D%28280.5-279.6%29%2F4.36%3D0.9%2F4.36%3D0.21%5C%5C%5C%5C%5C%5CP%28X%3D280%29%3DP%28-0.02%3Cz%3C0.21%29%3DP%28z%3C0.21%29-P%28z%3C-0.02%29%5C%5C%5C%5CP%28X%3D280%29%3D0.583-0.492%3D0.091)