Question

HELP ASAP!

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 229.5-cm and a standard deviation of 2.3-cm. For shipment, 29 steel rods are bundled together. Find the probability that the average length of a randomly selected bundle of steel rods is between 229.9-cm and 230.4- cm. P(229.9-cm <M<230.4-cm) - Enter your answer as a number accurate to 4 decimal places. Ansters obtained using exact z-scores or 2-scores rounded to 3 decimal places are accepted. Dante

Answer 1

Given :

Mean = μ = 229.5 cm

Standard deviation = σ = 2.3 cm

n = 29 steel rods

Solution :

This can be solved using central limit theorem,

229.9 - - 230.4 - u P(229.9cm <M < 230.4cm) =P <<<

  $= P\left [ \frac{229.9-229.5 }{\frac{2.3 }{\sqrt{29}}} < Z < \frac{230.4-229.5 }{\frac{2.3 }{\sqrt{29}}}\right ]$

  $= P\left [ \frac{0.4 }{0.427} < Z < \frac{0.9}{0.427}\right ]$

$= P\left [ 0.94 < Z < 2.11\right ]$

  

=1-PZ > 0.94 - PZ > 2.11

Using normal probability integral table we will get,

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