Assume the random variable x is normally distributed with mean mu equals 50μ=50 and standard deviation sigma equals 7σ=7. compute the probability. be sure to draw a normal curve with the area corresponding to the probability shaded. upper p left parenthesis upper x greater than 41 right parenthesisp(x>41) which of the follow

Answer:[tex]P(34<X<63)=P(\frac{34-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{63-\mu}{\sigma})=P(\frac{34-50}{7}<Z<\frac{63-50}{7})=P(-2.286<z<1.857)[/tex] And we can find this probability with this difference and using the normal standard table or excel: [tex]P(-2.286<z<1.857)=P(z<1.857)-P(z<-2.286)=0.968-0.0111= 0.9569[/tex] And the result is illustrated in the figure attached.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".  Solution to the problem Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by: [tex]X \sim N(50,7)[/tex]  Where [tex]\mu=50[/tex] and [tex]\sigma=7[/tex] We are interested on this probability [tex]P(34<X<63)[/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(34<X<63)=P(\frac{34-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{63-\mu}{\sigma})=P(\frac{34-50}{7}<Z<\frac{63-50}{7})=P(-2.286<z<1.857)[/tex] And we can find this probability with this difference and using the normal standard table or excel: [tex]P(-2.286<z<1.857)=P(z<1.857)-P(z<-2.286)=0.968-0.0111= 0.9569[/tex] And the result is illustrated in the figure attached.