A company produces steel rods.

Question

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 172.1-cm and a standard deviation of 1.5-cm. For shipment, 26 steel rods are bundled together.

Find the probability that the average length of a randomly selected bundle of steel rods is greater than 172.5-cm.

P(M > 172.5-cm) = 

Enter your answer as a number accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

Answer 1

The new random variable is the average length of the bundle.

Y=(X1+X2+....X26)/26

Mean: E[Y]=E[X1+X2+....X2626]=172.1

Variance: V[Y]=V[X1+X2+....X2626]=1262∗(V[X1+X2+....X26])=26∗1.52262=1.5226=0.086538

SD = (√Variance)=0.294174

Now to find the number in Z,

P(Y>172.5)

= P(Y−172.10.294>172.5−172.10.294)

= P(Z>1.36)

= 1−P(Z<1.36) #We can use z table to find this number.

= 1−Φ(1.36)

= 1−0.91309

= 0.08691

The video has more details

Answer 2

Hi Ashley,

For this problem, we will be using the Z-score formula where:

Z = X-bar (value) - u (population mean) / (SD/n) , where SD = standard deviation and n = sample size of steel rods

We need to first solve for the Z-score

Z = X-bar (value) - u (population mean) / (SD / sqrt(n))

Z = (172.5 - 172.1) / (1.5 / sqrt(26))

Z = 1.36

Now, this problem was initially to find P(x > 172.5 cm) but now it's P(Z > 1.36) which we will find from a Z-table. From the Z-table, we get 0.9131.

Hope this helps!