Respuesta :
Answer:
a) [tex]P(64<X<78)=P(\frac{64-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{78-\mu}{\sigma})=P(\frac{64-71}{3}<Z<\frac{78-71}{3})=P(-2.33<Z<2.33)[/tex]
[tex]P(-2.33<Z<2.33)=P(Z<2.33)-P(Z<-2.33)[/tex]
[tex]P(-2.33<Z<2.33)=P(Z<2.33)-P(Z<-2.33)=0.9901-0.0099=0.9802[/tex]
b) [tex] 1.96 = \frac{(71+c) -71}{3}[/tex]
And solving for c we got:
[tex] 1.96 * 3= c , c= 5.88[/tex]
So then the value of c = 5.88
c) Let X the niumber of acceptable specimens among a sampel of 10. The acceptable range from part a) is between 64 and 78 and we found that [tex] P(64<X<78) = 0.9802[/tex] that would represent our p value, and we select samples of size n =10, so then the expected value would be:
[tex] E(X)= n*p = 10*0.9802= 9.8[/tex]
d) [tex] P(Y \leq 8) = 1-P(Y>8) = 1-[P(Y\geq 9)]= 1-[P(Y=9)+P(Y=10)][/tex]
[tex]P(Y=9)=(10C9)(0.8)^9 (1-0.8)^{10-9}=0.2684[/tex]
[tex]P(Y=10)=(10C10)(0.8)^{10} (1-0.8)^{10-10}=0.1074[/tex]
[tex] P(Y \leq 8) = 1-P(Y>8) = 1-[P(Y\geq 9)]= 1-[P(Y=9)+P(Y=10)] = 1-(0.2684+0.1074)= 0.6242[/tex]
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Part a
Let X the random variable that represent the hardness of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(71,3)[/tex]
Where [tex]\mu=71[/tex] and [tex]\sigma=3[/tex]
We are interested on this probability
[tex]P(64<X<78)[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
If we apply this formula to our probability we got this:
[tex]P(64<X<78)=P(\frac{64-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{78-\mu}{\sigma})=P(\frac{64-71}{3}<Z<\frac{78-71}{3})=P(-2.33<Z<2.33)[/tex]
And we can find this probability with this difference:
[tex]P(-2.33<Z<2.33)=P(Z<2.33)-P(Z<-2.33)[/tex]
And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.
[tex]P(-2.33<Z<2.33)=P(Z<2.33)-P(Z<-2.33)=0.9901-0.0099=0.9802[/tex]
Part b
For this case we want to find the value of c who satisfy this condition:
[tex] P(71-c < X< 71+c) = 0.95[/tex]
For this case we can use the z score formula, we can find two values on the normal dtandard distribution that accumulates 0.95 of the values on the middle and for this case the two values are:
[tex] z_{crit}= \pm 1.96[tex]
The reason is because [tex] P(-1.96 <Z<1.96)=0.95[/tex]
So using the z score formula we have this:
[tex] -1.96 = \frac{(71-c) -71}{3}[/tex]
And solving for c we got:
[tex] -1.96 * 3= -c , c= 5.88[/tex]
[tex] 1.96 = \frac{(71+c) -71}{3}[/tex]
And solving for c we got:
[tex] 1.96 * 3= c , c= 5.88[/tex]
So then the value of c = 5.88
Part c
Let X the niumber of acceptable specimens among a sampel of 10. The acceptable range from part a) is between 64 and 78 and we found that [tex] P(64<X<78) = 0.9802[/tex] that would represent our p value, and we select samples of size n =10, so then the expected value would be:
[tex] E(X)= n*p = 10*0.9802= 9.8[/tex]
Part d
For this case we define the random variable Y= the number among the ten specimens with hardness less than 73.52, for this case we can find the probability like this:
[te]x P(X<73.52) = P(Z< \frac{73.52-71}{3}=P(Z<0.84)=0.800[/tex]
So then we can define our random variable like this:
[tex] Y \sim Bin (n=10, p =0.8)[/tex]
And we want to find this probability: [tex] P(Y \leq 8)[/tex]
The probability mass function for the Binomial distribution is given as:
[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]
For this case we can use the complement rule:
[tex] P(Y \leq 8) = 1-P(Y>8) = 1-[P(Y\geq 9)]= 1-[P(Y=9)+P(Y=10)][/tex]
[tex]P(Y=9)=(10C9)(0.8)^9 (1-0.8)^{10-9}=0.2684[/tex]
[tex]P(Y=10)=(10C10)(0.8)^{10} (1-0.8)^{10-10}=0.1074[/tex]
[tex] P(Y \leq 8) = 1-P(Y>8) = 1-[P(Y\geq 9)]= 1-[P(Y=9)+P(Y=10)] = 1-(0.2684+0.1074)= 0.6242[/tex]
