Respuesta :
Answer:
The probability that the the total length after insertion is between 34.5 and 35 inches is 0.1589.
Step-by-step explanation:
Let the random variable X represent the length of the first piece, Y represent the length of the second piece and Z represents the overlap.
It is provided that:
[tex]X\sim N(20,\ 0.50^{2})\\Y\sim N(15,\ 0.40^{2})\\Z\sim N(1,\ 0.10^{2})[/tex]
It is provided that the lengths and amount of overlap are independent of each other.
Compute the mean and standard deviation of total length as follows:
[tex]E(T)=E(X+Y-Z)\\=E(X)+E(Y)-E(Z)\\=20+15-1\\=34[/tex]
[tex]SD(T)=\sqrt{V(X+Y-Z)}\\=\sqrt{V(X)+V(Y)+V(Z)}\\=\sqrt{0.50^{2}+0.40^{2}+0.10^{2}}\\=0.6480741\\\approx 0.65[/tex]
Since X, Y and Z all follow a Normal distribution, the random variable T, representing the total length will also follow a normal distribution.
[tex]T\sim N(34, 0.65^{2})[/tex]
Compute the probability that the the total length after insertion is between 34.5 and 35 inches as follows:
[tex]P(34.5<T<35)=P(\frac{34.5-34}{0.65}<\frac{T-\mu_{T}}{\sigma_{T}}<\frac{35-34}{0.65})\\\\=P(0.77<Z<1.54)\\\\=P(Z<1.54)-P(Z<0.77)\\\\=0.93822-0.77935\\\\=0.15887\\\\\approx 0.1589[/tex]
*Use a z-table.
Thus, the probability that the the total length after insertion is between 34.5 and 35 inches is 0.1589.
