Mathematics, 06.05.2020 03:19 tynasiaparks13

The weights of a certain brand of candies are normally distributed with a mean weight of 0.8556 g and a standard deviation of 0.0511 g. A sample of these candies came from a package containing 469 candies, and the package label stated that the net weight is 400.3 g. (If every package has 469 candies, the mean weight of the candies must exceed StartFraction 400.3 Over 469 EndFraction equals0.8536 g for the net contents to weigh at least 400.3 g.)

Required:
a. If 1 candy is randomly selected, find the probability that it weighs more than 0.8536 g.
b. If 441 candies are randomly selected, find the probability that their mean weight is at least 0.8543 g.
c. Given these results, does it seem that the candy company is providing consumers with the amount claimed on the label?

Answers: 1

Show answers

Answers

Answer from: mtammy884

a) There is a probability P=0.4992 that a randomly selected candy weights at least 0.8543 g.

b) There is a probability P=0.3712 that a randomly selected sample of 441 candies have an average weight of at least 0.8543 g.

c) The claim of the brand should be re-evaluated as it is not providing consumers with the amount claimed on the label.

Step-by-step explanation:

The average weight of the candies in the package is:

$\bar x=\dfrac{\sum x_i}{n}=\dfrac{400.3}{469}=0.8535$

For a randomly selected candy (is a sample of size n=1) the standard deviation is the population standard deviation (sigma=0.0511 g).

The probabiltiy that is weights more than 0.8536 can be calculated with the z-score.

$z=\dfrac{X-\mu}{\sigma/\sqrt{n}}= \dfrac{0.8536-0.8535}{0.0511/\sqrt{1}}=\dfrac{0.0001}{0.0511}=0.002$

P(X0.8536)=P(z0.002)=0.4992

There is a probability P=0.4992 that a randomly selected candy weights at least 0.8536 g.

b. If the sample is now of n=441 candies, and we want to know the probability that the mean weight is at least 0.8543 g, the z-score needs to be recalculated:

$z=\dfrac{X-\mu}{\sigma/\sqrt{n}}=\dfrac{0.8543-0.8535}{0.0511/\sqrt{441}}=\dfrac{0.0008}{0.0024}=0.3288$

P(X_s0.8543)=P(z0.3288)=0.37115

There is a probability P=0.3712 that a randomly selected sample of 441 candies have an average weight of at least 0.8543 g.

c. To be more confident about the claim that the mean weight is 0.8556 g, we can calculate the probability that, for a package of 469 candies and using the mean of 0.8535 g that we calculated before, the average weight is at least 0.8556 g.

$z=\dfrac{X-\mu}{\sigma/\sqrt{n}}=\dfrac{0.8556-0.8535}{0.0511/\sqrt{469}}=\dfrac{0.0021}{0.0024}=0.89$

P(X_s0.8556)=P(z0.89)=0.18673

The probability is P=0.187, so the claim of the brand should be re-evaluated as it is not providing consumers with the amount claimed on the label.

Answer from: Quest

$450

step-by-step explanation:

first, you have to multiply 150x2=300

next, you have to add 150+300= 450

Answer from: Quest

The academic affairs council is the highest expert body processing academic affairs, as stipulated by the university statutes. specialist aaro häkkinen (educational policy) took part in the council’s meeting. the agenda of the meeting included bachelors’ feedback, the timing of pedagogical studies, the grant system for tuition fees, the standardisation of administrative practices for internships and the situation of foreign-language and multilingual master’s programmes.

Another question on Mathematics

Mathematics, 21.06.2019 15:30

Given the box plot, will the mean or the median provide a better description of the center? box plot with min at 10.5, q1 at 11.5, median at 12.5, q3 at 13.5, max at 15

Answers: 2

Answer

Mathematics, 21.06.2019 20:00

Which expression is equivalent to -5(3x - 6/7)

Answers: 1

Answer

Mathematics, 21.06.2019 21:00

Timmy uses 1 1 2 cups of sugar for every 2 gallons of lemonade he makes. how many cups of sugar would timmy use if he makes 13 gallons of lemonade?

Answers: 1

Answer

Mathematics, 21.06.2019 22:10

Monitors manufactured by tsi electronics have life spans that have a normal distribution with a standard deviation of 1800 hours and a mean life span of 20,000 hours. if a monitor is selected at random, find the probability that the life span of the monitor will be more than 17,659 hours. round your answer to four decimal places.

Answers: 2

Answer

You know the right answer?

The weights of a certain brand of candies are normally distributed with a mean weight of 0.8556 g an...

Questions