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.
Answer:
x = 5
Step-by-step explanation:
5x + 17 = 42
5x = 25
x = 5
The answer is B because the mean is 131.6 and the median is 36, and obviously 131.6 is the greater value. Hope this helps! If I am wrong, someone in the comments please correct me
Answer:
Step-by-step explanation:
At the first place is better because
$0.8 * 5 = $ 4
$4.5 / 5 = $0.9
Is more expensive in the another place