Find the area of the semicircle. Round your answer to the nearest hundredth.
1 answer:
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Answer:
The probability that the average length of rods in a randomly selected bundle of steel rods is greater than 259 cm is 0.65173.
Step-by-step explanation:
We are given that a company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 259.2 cm and a standard deviation of 2.1 cm. For shipment, 17 steel rods are bundled together.
Let = <u><em>the average length of rods in a randomly selected bundle of steel rods</em></u>
The z-score probability distribution for the sample mean is given by;
Z = ~ N(0,1)
where, = population mean length of rods = 259.2 cm
= standard deviaton = 2.1 cm
n = sample of steel rods = 17
Now, the probability that the average length of rods in a randomly selected bundle of steel rods is greater than 259 cm is given by = P( > 259 cm)
P( > 259 cm) = P(
>
) = P(Z > -0.39) = P(Z < 0.39)
= <u>0.65173</u>
The above probability is calculated by looking at the value of x = 0.39 in the z table which has an area of 0.65173.