Respuesta :
Answer:
The probability is 0.6826
Step-by-step explanation:
Lets call X the distribution of the length of the fish. X has distribution N(μ,σ). We can standarize X to obtain a random variable N(0,1)
[tex] W = \frac{X-\mu}{\sigma} \simeq N(0,1) [/tex]
We can use this standarization to find the probability we are looking for. The values of the cumulative distribution function of the standar Normal, lets call it [tex] \phi [/tex] , are tabulated; you can find the values of that function in the attached file.
[tex]P(\mu-\sigma < X < \mu+\sigma) = P(\frac{\mu-\sigma-\mu}{\sigma} < \frac{X-\mu}{\sigma} < \frac{\mu+\sigma-\mu}{\sigma}) = P(-1<W<1)[/tex]
The tabulated values of [tex] \phi [/tex] are only for positive numbers. We can use the symmetry of the density function of the normal distribution to continue the computation
[tex] P(-1<W<1) = P(W<1) - P(W<-1) = \phi(1)-\phi(-1) =\\
\phi(1) - (1-\phi(1)) = 2\phi(1) -1 = 2*0.8413-1 = 0.6826[/tex]
Thus, the probability is 0.6826.
