During the baseball season, marvins team won 5 games and lost 14 games

Mathematics

1 answer:

777dan777 [17]3 years ago

7 0

The answer is 9 games

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A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 259.2-cm and a standard dev

Varvara68 [4.7K]

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 $\bar X$ = the average length of rods in a randomly selected bundle of steel rods

The z-score probability distribution for the sample mean is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean length of rods = 259.2 cm

$\sigma$ = 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( $\bar X$ > 259 cm)

P( $\bar X$ > 259 cm) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{259-259.2}{\frac{2.1}{\sqrt{17} } }$ ) = P(Z > -0.39) = P(Z < 0.39)