A box of breakfast cereal has "contains 500g of breakfast cereal" printed on it. Suppose that in fact the weight of breakfast cereal contained in these boxes is normally distributed with a mean of 512g and a st. deviation of 8g.

Determine the probability that a randomly chosen box of cereal contains less then 500g

Question

A box of breakfast cereal has "contains 500g of breakfast cereal" printed on it. Suppose that in fact the weight of breakfast cereal contained in these boxes is normally distributed with a mean of 512g and a st. deviation of 8g.

Determine the probability that a randomly chosen box of cereal contains less then 500g

kropot72 · Answer

This can be solved by finding the z-score for 500g and then using a standard normal distribution table. Do you need help with these?

kropot72 · Answer

The z-score can be found from the formula:
$$\large z=\frac{X-\mu}{\sigma}$$

rane · Answer

X = the given weight or?

kropot72 · Answer

X is the given value of the random variable. In this case X = 500g.

kropot72 · Answer

$$\large z=\frac{X-\mu}{\sigma}=\frac{500-512}{8}=you\ can\ calculate$$
After you calculate the z-score, you need to use a standard normal distribution table to find the required probability.

rane · Answer

1.5

perl · Answer

you can use a calculator TI 83 or TI 84

perl · Answer

normalcdf ( -1E99, 500, 512, 8 ) =

kropot72 · Answer

Not really. What is the value of the numerator:
500 - 512 = ?

perl · Answer

.0668

rane · Answer

500-512 = -12/8 = -1.5

perl · Answer

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