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A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 117.7-cm and a standard deviation of 2.2-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 greater than 118.5-cm.

Respuesta :

Answer:

Required probability is 0.9748

Step-by-step explanation:

given data

mean [tex]\mu[/tex] = 117.7-cm

standard deviation [tex]\sigma[/tex] = 2.2-cm

sample size n = 29

solution

we consider here random variable which represents here length of rod= x

so get here first z that is express as

[tex]Z = \dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]    

put here value with x value 118.5-cm

[tex]Z = \dfrac{118.5-117.7}{\dfrac{2.2}{\sqrt{29}}}[/tex]

Z = 1.9582

p value is 0.9748

so required probability is 0.9748

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