Respuesta :
Answer:
11.69% probability that it will weigh between 523 grams and 534 grams
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 551, \sigma = 20[/tex]
If you pick one fruit at random, what is the probability that it will weigh between 523 grams and 534 grams
This is the pvalue of Z when X = 534 subtracted by the pvalue of Z when X = 523. So
X = 534
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{534 - 551}{20}[/tex]
[tex]Z = -0.85[/tex]
[tex]Z = -0.85[/tex] has a pvalue of 0.1977
X = 523
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{523 - 551}{20}[/tex]
[tex]Z = -1.4[/tex]
[tex]Z = -1.4[/tex] has a pvalue of 0.0808
0.1977 - 0.0808 = 0.1169
11.69% probability that it will weigh between 523 grams and 534 grams
Answer:
[tex]P(523<X<534)=P(\frac{523-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{534-\mu}{\sigma})=P(\frac{523-551}{20}<Z<\frac{534-551}{20})=P(-1.4<z<-0.85)[/tex]
And we can find this probability with this difference:
[tex]P(-1.4<z<-0.85)=P(z<-0.85)-P(z<-1.4)[/tex]
And in order to find these probabilities we can use the table for the normal standard distribution, excel or a calculator.
[tex]P(-1.4<z<-0.85)=P(z<-0.85)-P(z<-1.4)=0.198-0.0808=0.1172[/tex]
Step-by-step explanation:
Let X the random variable that represent the weigths of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(551,20)[/tex]
Where [tex]\mu=551[/tex] and [tex]\sigma=20[/tex]
We are interested on this probability
[tex]P(523<X<534)[/tex]
We can solve the 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(523<X<534)=P(\frac{523-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{534-\mu}{\sigma})=P(\frac{523-551}{20}<Z<\frac{534-551}{20})=P(-1.4<z<-0.85)[/tex]
And we can find this probability with this difference:
[tex]P(-1.4<z<-0.85)=P(z<-0.85)-P(z<-1.4)[/tex]
And in order to find these probabilities we can use the table for the normal standard distribution, excel or a calculator.
[tex]P(-1.4<z<-0.85)=P(z<-0.85)-P(z<-1.4)=0.198-0.0808=0.1172[/tex]
