Question

The weights of a certain brand of candies are normally distributed with a mean weight of 0.8552 g and a standard deviation of 0.0519 g. A sample of these candies came from a package containing 442 ​candies, and the package label stated that the net weight is 377.3 g.​ (If every package has 442 ​candies, the mean weight of the candies must exceed StartFraction 377.3 Over 442 EndFraction equals0.8537 g for the net contents to weigh at least 377.3 ​g.) a. If 1 candy is randomly​ selected, find the probability that it weighs more than 0.8537 g. The probability is nothing. ​(Round to four decimal places as​ needed.)

Answer 1

Answer:

The probability that the weight of a candy randomly selected is more than 0.8537 is 0.7486

Step-by-step explanation:

The given parameters are;

The mean candle weight = 0.8552 g

The standard deviation = 0.0519 g

The number in the sample, n = 442 candles

By central limit theorem, the sample standard deviation, $\sigma _{\bar x}$ is given by the relationship;

$\sigma _{\bar x} = \dfrac{\sigma}{\sqrt{n} } = \dfrac{0.0519}{\sqrt{442} } = 0.002469$

The probability is given by the relation;

$P\left (\bar{X}>0.8537 \right )= P\left (\dfrac{\bar{X}-\mu }{\dfrac{\sigma }{\sqrt{n}}} >\dfrac{0.8537-\mu }{\dfrac{\sigma }{\sqrt{n}}} \right )$

$P\left (\bar{X}>0.8537 \right )= P\left (\dfrac{\bar{X}-0.8552 }{\dfrac{\sigma }{\sqrt{n}}} >\dfrac{0.8537-0.8552 }{\dfrac{0.0519 }{\sqrt{442}}} \right )$

$P\left (\bar{X}>0.8537 \right )= P\left (z>-0.6076\right )$

The from the z-score table we have = 0.2514

The probability of P (z > -6076) = 1 - 0.2514 =  0.7486

The probability that the weight of a candy randomly selected is more than 0.8537 = 0.7486.