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Vlada [557]
3 years ago
5

Based on a poll 62% of internet users are more careful about personal information when using a public wifi hotspot. What is the

probability that amoung three randomly selected internet users at least one is more careful about personal information when using pubblic wifi hotspot?
Mathematics
1 answer:
Setler79 [48]3 years ago
7 0

Answer:

0.9451

Step-by-step explanation:

Remaining question? <em>"How is the result affected by the additional information that the survey subjects volunteered to​respond?"</em>

Probability that at least 1 user is more careful about personal information when using a public Wi-Fi hot spot is:

P(X≥1) = 1 − P(X<1)

= 1 − P(X=0)

= 1 - [(3,0) (0.62)^0 (1-0.62)^3-0

= 1 - 0.054872

= 0.945128

= 0.9451

Thus, the probability that among three randomly selected Internet users; at least one is more careful about personal information when using a public Wi-Fi hotspot is 0.9451

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Zen spent $255 on a bag and a belt. She wanted to buy another
Galina-37 [17]

Answer:

A)$ 45

B) $105

Step-by-step explanation:

Bag and a belt cost $255

Let bag = x

Let belt = y

X+y= 255 equation 1

Let total money be z first

Remaining money= z-255

X-30 = z-255

Y +15 = z-255

Equating the left side of the equation

X+30 = y+15

X-y= 45 equation 2

Solving simultaneously

X+y= 255

X-y= 45

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150-y= 45

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3 years ago
The concentration of particles in a suspension is 50 per mL. A 5 mL volume of the suspension is withdrawn. a. What is the probab
kolezko [41]

Answer:

(a) 0.6579

(b) 0.2961

(c) 0.3108

(d) 240

Step-by-step explanation:

The random variable <em>X</em> can be defined as the number of particles in a suspension.

The concentration of particles in a suspension is 50 per ml.

Then in 5 mL volume of the suspension the number of particles will be,

5 × 50 = 250.

The random variable <em>X</em> thus follows a Poisson distribution with parameter, <em>λ</em> = 250.

The Poisson distribution with parameter λ, can be approximated by the Normal distribution, when λ is large say λ > 10.  

The mean of the approximated distribution of X is:

μ = λ  = 250

The standard deviation of the approximated distribution of X is:

σ = √λ  = √250 = 15.8114

Thus, X\sim N(250, 250)

(a)

Compute the probability that the number of particles withdrawn will be between 235 and 265 as follows:

P(235

                             =P(-0.95

Thus, the value of P (235 < <em>X</em> < 265) = 0.6579.

(b)

Compute the probability that the average number of particles per mL in the withdrawn sample is between 48 and 52 as follows:

P(48

                             =P(-0.28

Thus, the value of P(48.

(c)

A 10 mL sample is withdrawn.

Compute the probability that the average number of particles per mL in the withdrawn sample is between 48 and 52 as follows:

P(48

                             =P(-0.40

Thus, the value of P(48.

(d)

Let the sample size be <em>n</em>.

P(48

                             0.95=P(-z

The value of <em>z</em> for this probability is,

<em>z</em> = 1.96

Compute the value of <em>n</em> as follows:

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Thus, the sample selected must be of size 240.

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