SOLUTION: A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 110.3 cm and a standard deviation of 1.7 cm. For shipment, 7 steel rods are bun

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Question 1206804: A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 110.3 cm and a standard deviation of 1.7 cm. For shipment, 7 steel rods are bundled together.
Find the probability that the average length of a randomly selected bundle of steel rods is more than 112 cm.
Found 2 solutions by MathLover1, math_tutor2020:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
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mean: mu=110.3cm
n=7 steel rods are bundled together
a standard deviation: sigma=1.7cm

find the standard error:
s+=+sigma%2Fsqrt%28n+%29=+1.7%2Fsqrt%287%29+%E2%89%88+0.6425

now find the z-score using the sample standard deviation to standardize the normal distribution:
z+=+%28mu-sigma%29%2Fs+=%28112-110.3%29%2F0.6425=2.6459
Using the standard normal distribution table or calculator, we can find the probability that the average length of a randomly selected bundle of steel rods is more than 112cm

P%28M%3E112%29+=+P%28Z%3E2.6459%29=0.00407369≈ 0.0041




Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

mu = 110.3 = population mean
sigma = 1.7 = population standard deviation

n = 7 = sample size
xbar = sample mean

Compute the z score when xbar = 112
z = (xbar - mu)/( sigma/sqrt(n) )
z = (112 - 110.3)/( 1.7/sqrt(7) )
z = 2.65 approximately when rounding to 2 decimal places

Then use a Z table to determine that
P(Z < 2.65) = 0.99598
which leads to
P(Z > 2.65) = 1-P(Z < 2.65)
P(Z > 2.65) = 1-0.99598
P(Z > 2.65) = 0.00402
This value is approximate.

To get more accuracy you can use a TI84 or similar stats calculator.