2527=3 3214=8 1249=4 7321=? A)7 B)3 C)9

A supermarket has two customers waiting to pay for their purchases at counter I and one customer waiting to pay at counter II. L

Pachacha [2.7K]

Answer:

b. 0.864

Step-by-step explanation:

Let's start defining the random variables.

Y1 : ''Number of customers who spend more than $50 on groceries at counter 1''

Y2 : ''Number of customers who spend more than $50 on groceries at counter 2''

If X is a binomial random variable, the probability function for X is :

$P(X=x)=(nCx)p^{x}(1-p)^{n-x}$

Where P(X=x) is the probability of the random variable X to assume the value x

nCx is the combinatorial number define as :

$nCx=\frac{n!}{x!(n-x)!}$

n is the number of independent Bernoulli experiments taking place

And p is the success probability.

In counter I :

Y1 ~ Bi (n,p)

Y1 ~ Bi(2,0.2)

$P(Y1=y1)=(2Cy1)(0.2)^{y1}(0.8)^{2-y1}$

With y1 ∈ {0,1,2}

And P( Y1 = y1 ) = 0 with y1 ∉ {0,1,2}

In counter II :

Y2 ~ Bi (n,p)

Y2 ~ Bi (1,0.3)

$P(Y2=y2)=(1Cy2)(0.3)^{y2}(0.7)^{1-y2}$

With y2 ∈ {0,1}

And P( Y2 = y2 ) = 0 with y2 ∉ {0,1}

(1Cy2) with y2 = 0 and y2 = 1 is equal to 1 so the probability function for Y2 is :

$P(Y2=y2)=(0.3)^{y2}(0.7)^{1-y2}$

Y1 and Y2 are independent so the joint probability distribution is the product of the Y1 probability function and the Y2 probability function.

$P(Y1=y1,Y2=y2)=P(Y1=y1).P(Y2=y2)$

$P(Y1=y1,Y2=y2)=(2Cy1)(0.2)^{y1}(0.8)^{2-y1}(0.3)^{y2}(0.7)^{1-y2}$

With y1 ∈ {0,1,2} and y2 ∈ {0,1}

P( Y1 = y1 , Y2 = y2) = 0 when y1 ∉ {0,1,2} or y2 ∉ {0,1}

b. Not more than one of three customers will spend more than $50 can mathematically be expressed as :

$Y1 + Y2 \leq 1$

$Y1 + Y2\leq 1$ when Y1 = 0 and Y2 = 0 , when Y1 = 1 and Y2 = 0 and finally when Y1 = 0 and Y2 = 1

To calculate $P(Y1+Y2\leq 1)$ we must sume all the probabilities that satisfy the equation :

$P(Y1+Y2\leq 1)=P(Y1=0,Y2=0)+P(Y1=1,Y2=0)+P(Y1=0,Y2=1)$

$P(Y1=0,Y2=0)=(2C0)(0.2)^{0}(0.8)^{2-0}(0.3)^{0}(0.7)^{1-0}=(0.8)^{2}(0.7)=0.448$

$P(Y1=1,Y2=0)=(2C1)(0.2)^{1}(0.8)^{2-1}(0.3)^{0}(0.7)^{1-0}=2(0.2)(0.8)(0.7)=0.224$

$P(Y1=0,Y2=1)=(2C0)(0.2)^{0}(0.8)^{2-0}(0.3)^{1}(0.7)^{1-1}=(0.8)^{2}(0.3)=0.192$

$P(Y1+Y2\leq 1)=0.448+0.224+0.192=0.864\\P(Y1+Y2\leq 1)=0.864$

7 0

2 years ago

How many solutions does the system have? { y = − 7 x + 8 y = − 7 x − 8 ⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ y=−7x+8 y=−7x−8

svp [43]

Answer:

woooo that is difficult

3 0

2 years ago

Read 2 more answers

weekly wages at a certain factory are normally distributed with a mean of $400 and a standard deviation of $50. find the probabi

kenny6666 [7]

Answer:

0.34134

Step-by-step explanation:

In other to solve for this question, we would be using the z score formula

z = (x - μ) / σ

x = raw score

μ = mean

σ = Standard deviation

We are told in the question to find the probability that a worker selected at random makes between $350 and $400

let x1 = 350 and x2= 400 with the mean μ = 400 and standard deviation σ = $50.

z1 = (x1 - μ) / σ = (350-400) / 50 = -1

z2 = (x2 - μ) / σ = (400 - 400) / 50 = (0/50) = 0

From tables, P(z <= -1) = 0.15866

P(z <= 0) = 0.5

Then, the probability would give us, P(-1 ≤ z ≤ 0) =0.5 - 0.15866 =

0.34134

Hence, The probability that a worker selected at random makes between $350 and $400 = 0.34134

7 0

2 years ago

Carol is going to buy a purse she found. It is originally $120.00, but has been marked down 25%. When Carol goes to buy the purs

Reil [10]

$82.80
I hope this helped!

4 0

2 years ago

Quadrilateral ABCD is inscribed in circle P as shown. Which statement is necessarily true? A. `"m"/_A +"m"/_B = "m"/_C + "m"/_D

pogonyaev

Opposite angles of an inscribed quadrilateral are supplementary, so it is true that ...

... B. m∠A + m∠C = m∠B + m∠D . . . . . = 180°

3 0

2 years ago

2527=33214=81249=47321=?A)7 B)3 C)9​

2527=33214=81249=47321=?A)7 B)3 C)9