contestada

Delicious Candy markets a two-pound box of assorted chocolates. Because of imperfections in the candy making equipment, the actual weight of the chocolate has a uniform distribution ranging from 31.8 to 32.6 ounces. a. Define a probability density function for the weight of the box of chocolate. b. What is the probability that a box weighs (1) exactly 32 ounces; (2) more than 32.3 ounces; (3) less than 31.8 ounces? c. The government requires that at least 60% of all products sold weigh at least as much as stated weight. Does Delicious Candy violate government regulation?

Respuesta :

opudodennis

Answer:

a. [tex]f(x)= 1.25\ \ \ \ , \ for \ 31.8\leq x\leq 32.6[/tex]

[tex]b. \ P(X=32)=0\\P(X>32.3)=0.375\\P(X<31.8)=0[/tex]

c. No.  Delicious Candy isn't violating any government regulations

Step-by-step explanation:

a.

-A uniform distribution is given by the formula:

[tex]f(x)=\frac{1}{b-a} \ \ \ for \ \ \ a\leq x\leq b[/tex]

#we substitute our values in the formula above to determine the distribution:

[tex]f(x)=\frac{1}{b-a}\\\\=\frac{1}{32.6-31.8}\\\\=1.25\\\\\therefore f(x)=1.25, \ \ \ 31.8\leq x\leq 32.6[/tex]

Hence, the probability density function for the box's weight is given as: [tex]f(x)=1.25, \ \ \ 31.8\leq x\leq 32.6[/tex]

b. The probability of the box's weight being exactly 32 ounces is obtained by integrating f(x) over a=b=32:

[tex]f(x)=1.25, \ \ \ a\leq x\leq b\\\\=\int\limits^{32}_{32} {1.25} \, dx \\\\\\=[1.25x]\limits^{32}_{32}\\\\\\=1.25[32.0-32.0]\\\\\\=0[/tex]

Hence,  the probability that a box weighs exactly 32 ounces is 0.000

ii.The probability that a box weighs more than 32.3 is obtained by integrating f(x) over the limits 32.3 to 32.6 :

[tex]f(x)=1.25, \ \ \ a\leq x\leq b\\\\=\int\limits^{32.6}_{32.3} {1.25} \, dx \\\\\\=[1.25x]\limits^{32.6}_{32.3}\\\\\\=1.25[32.6-32.3]\\\\\\=0.375[/tex]

Hence, the probability that a box weighs more than 32.3 ounces is 0.3750

iii. The probability that a box weighs less than 31.8 is 0.000 since the weight limits are [tex]31.8\leq x\leq 32.6[/tex].

-Any value above or below these limits have a probability of 0.000

c. Let 32 ounces be the government's stated weight.

[tex]1.25(32.6-32)=0.75\\\\0.75>0.60[/tex]

Hence, Delicious Candy isn't violating any government's regulations.

(a): The required probability density function for the weight of the box of chocolate is 1.25

(b):  The probability that a box weighs (1) exactly 32 ounces is 0

and  (2) more than 32.3 ounces is 0.375

(c): Therefore, Delicious Candy does not violate government regulation.

Probability:

Probability is a measure of the likelihood of an event to occur. Many events cannot be predicted with total certainty. We can predict only the chance of an event to occur i.e. how likely they are to happen, using it. Probability can range in from 0 to 1.

Given that,

The uniform distribution between [tex]a = 31.8[/tex] ounce and [tex]b = 32.6[/tex] ounce

Part(a):

The probability density function for the weight of the box of chocolate is,

[tex]\frac{1}{b-a}=\frac{1}{32.6-31.8} \\=1.25[/tex]

Part(b):

(1) P(exactly 32 ounces) = 0, because this is a continuous distribution.

(2) P(more than 32.3 ounces) =[tex]1.25\times (32.6-32.3)=0.375[/tex]

Part(c):

The stated weight of Delicious Candy =  2 pounds

That is, [tex]2\times 16=32[/tex] ounces

P(a candy weigh at least as much as stated) = P(at least 32)

[tex]1.25\times (32.6-32)=0.75[/tex]

So, 75% of candies weigh at least as much as stated.

Learn more about the topic of Probability:

https://brainly.com/question/6649771