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svp [43]
2 years ago
13

Company F sells fabrics known as fat quarters, which are rectangles of fabric created by cutting a yard of fabric into four piec

es. Occasionally the manufacturing process results in a fabric defect. Let the random variable X represent the number of defects on a fat quarter created by Company F. The following table shows the probability distribution of X.
X 0 1 2 3 4 or more
probability 0.58 0.23 0.11 0.05 0.03
If a fat quarter has more than 2 defects, it cannot be sold and is discarded. Let the random variable Y represent the number of defects on a fat quarter that can be sold by Company F.

(a) Construct the probability distribution of the random variable Y.

(b) Determine the mean and standard deviation of Y. Show your work.

Company G also sells fat quarters. The mean and standard deviation of the number of defects on a fat quarter that can be sold by Company G are 0.40 and 0.66, respectively. The fat quarters sell for $5.00 each but are discounted by $1.50 for each defect found.

(c) What are the mean and standard deviation of the selling price for the fat quarters sold by Company G?
Mathematics
1 answer:
jeka942 years ago
7 0

Answer:

a) Y 0 1 2

P(Y) 0.58 0.23 0.11

b) mean= 0.45, S.D= 0.6718

c) mean= 1.285, S.D= 8.74

Step-by-step explanation:

a) The following table shows the probability distribution of X:

X 0 1 2 3 4 or more

P(X) 0.58 0.23 0.11 0.05 0.03

Defect >2 = cannot be sold

Y = the number of defects on a fat quarter that can be sold by Company F.

Y = defect that can be sold

Y = Defect less or equal to 2 = 0,1,2

Probability distribution of the random variable Y:

Y 0 1 2

P(Y) 0.58 0.23 0.11

b) mean of Y (μ)

μ = Σ x*P(Y)

= (0*0.58) +(1*0.23)+(2*0.11)

= 0+0.23+0.22 = 0.45

Standard deviation of Y = σ

σ = Σ√(x-mean)^2*P(Y)

= Σ√[(x- μ )^2*P(Y)]

= √[(0-0.45)^2*0.58+ (1-0.45)^2*0.23 + (2-0.45)^2*0.11]

= √[0.11745 + 0.069575 +0.264275

= √(0.4513

σ = 0.6718

Company G:

σ for defect that be sold = 0.66

μ for defect that be sold = 0.40

Difference between μ of F and μ of G

= 0.45-0.40 = 0.05

Difference between σ of F and σ of G

= 0.67-0.66 = 0.01

Selling price of fat quarter without defect = $5

Discount per defect = $1.5

Selling price per defect = 5-1.5 = $3.5

Discount per 2 defect = $1.5*2 = $3

Selling price per defect = 5-3 = $2

Since defect to be sold cannot be greater than 2, let Y = 5,3,2

Probability distribution of the selling price Y:

Y 5 3 2

P(Y) 0.58 0.23 0.11

μ = (5*0.58) +(3.5*0.23)+(2*0.11)

μ = 2.9+0.805+0.22 =1.285

σ = Σ√[(x- μ )^2*P(Y)]

σ = √[(5-1.285)^2*0.58+ (3-1.285)^2*0.23 + (2-1.285)^2*0.11]

σ = 8.00+0.68+0.06 = 8.74

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Answer:

The  value is  P(A) =  0.133617

Step-by-step explanation:

From the question we are told that

  The  mean is  \mu =  7.5

  The standard deviation is  \sigma  =  0.2

  The safest water level is  between  7.2 and  7.8

Generally the probability that the selected pool has a pH level that is not considered safe is mathematically represented as

       P(A) =  1 - P(7.2 \le X  \le 7.8 )

Here  

      P(7.2 < X  < 7.8 ) = P(\frac{ 7.2 - \mu }{\sigma } <  \frac{X - \mu }{ \sigma }

Generally \frac{X - \mu }{ \sigma } =  Z (The  \ standardized \  value  \  of  X )

So

 P(7.2 < X  < 7.8 ) = P(\frac{ 7.2 - 7.5 }{0.2 } < Z    

 P(7.2 < X  < 7.8 ) = P(-1.5 < Z  

=>  P(7.2 < X  < 7.8 ) = P(Z <  1.5) - P(  Z < - 1.5)

From the z-table the probability of  (Z <  -1.5) and   (  Z <1.5)  are

    P(Z <  1.5) =  0.93319

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    P(Z <  -1.5) =  0.066807

So

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5 0
1 year ago
A 12-centimeter rod is held between a flashlight and a wall as shown. Find the length of the shadow on
choli [55]

Answer:

48 cm

Step-by-step explanation:

Given:

Distance of rod from the wall = 45 cm

Distance of rod from the light = 15 cm

Length of rod = 12 cm

We can see that <DAM and <BAF are equal

Also, <DMA and <BFM are equal because they are corresponding angles

To find the length of the shadow, let's take the equation

\frac{DM}{BF} = \frac{AM}{A.F}

Where.:

DM = ½ of length of the rod = ½*12 = 6

A.F = 15 + 45 = 60 cm

AM = 15 cm

Therefore,

\frac{DM}{BF} = \frac{AM}{A.F}

= \frac{6}{BF} = \frac{15}{60}

Cross multiplying, we have:

15 * B.F = 60 * 6

15 * B.F = 360

BF = \frac{360}{15}

BF = 24 cm

The shadow on the wall =

2 * BF

= 2 * 24

= 48 cm

The shadow on the wall is 48 cm

7 0
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