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Aleksandr-060686 [28]

3 years ago

12

The number of clients arriving at a bank machine is Poisson distributed with an average of 2 per minute. What is the probability

that no more than 2 customers will arrive in a minute

1 answer:

IrinaVladis [17]3 years ago

6 0

Answer:

The probability that no more than '2' customers will arrive in minute = 1.5412

Step-by-step explanation:

Explanation:-

Mean of the Poisson distribution 'λ' = 2 per minute

P(X=x) = e⁻ˣ λˣ/x!

The probability that no more than '2' customers will arrive in minute

P(x≤ 2) = P(x=0) +P(x=1)+P(x=2)

= e⁻² (2)°/0! + e⁻²(2)¹/1!+e⁻² (2)²/2!

= 1 + 0.2706 + 0.2706

= 1.5412

The probability that no more than '2' customers will arrive in minute = 1.5412

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A grocery store purchases melons from two distributors, J & K. Distributor J provides melons from organic farms. The distrib

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Answer: green.

Step-by-step explanation:

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3 years ago

Read 2 more answers

In a paint factory, an old conveyer line has filled 50 barrels of paint, and is filling more at a rate of 2 barrels per minute.

bezimeni [28]

Answer:

60 barrels

Step-by-step explanation:

y = # of barrels; x = minutes

worker: y = 12x

old line: y = 2x + 50

12x = 2x + 50

10x = 50

x = 5

y = 12(5) = 60

y = 2(5) + 50 = 60

60 barrels

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2 years ago

The following results come from two independent random samples taken of two populations.

photoshop1234 [79]

Answer:

$(a)\ \bar x_1 - \bar x_2 = 2.0$

$(b)\ CI =(1.0542,2.9458)$

$(c)\ CI = (0.8730,2.1270)$

Step-by-step explanation:

Given

$n_1 = 60$ $n_2 = 35$

$\bar x_1 = 13.6$ $\bar x_2 = 11.6$

$\sigma_1 = 2.1$ $\sigma_2 = 3$

Solving (a): Point estimate of difference of mean

This is calculated as: $\bar x_1 - \bar x_2$

$\bar x_1 - \bar x_2 = 13.6 - 11.6$

$\bar x_1 - \bar x_2 = 2.0$

Solving (b): 90% confidence interval

We have:

$c = 90\%$

$c = 0.90$

Confidence level is: $1 - \alpha$

$1 - \alpha = c$

$1 - \alpha = 0.90$

$\alpha = 0.10$

Calculate $z_{\alpha/2}$

$z_{\alpha/2} = z_{0.10/2}$

$z_{\alpha/2} = z_{0.05}$

The z score is:

$z_{\alpha/2} = z_{0.05} =1.645$

The endpoints of the confidence level is:

$(\bar x_1 - \bar x_2) \± z_{\alpha/2} * \sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}$

$2.0 \± 1.645 * \sqrt{\frac{2.1^2}{60}+\frac{3^2}{35}}$

$2.0 \± 1.645 * \sqrt{\frac{4.41}{60}+\frac{9}{35}}$

$2.0 \± 1.645 * \sqrt{0.0735+0.2571}$

$2.0 \± 1.645 * \sqrt{0.3306}$

$2.0 \± 0.9458$

Split

$(2.0 - 0.9458) \to (2.0 + 0.9458)$

$(1.0542) \to (2.9458)$

Hence, the 90% confidence interval is:

$CI =(1.0542,2.9458)$

Solving (c): 95% confidence interval

We have:

$c = 95\%$

$c = 0.95$

Confidence level is: $1 - \alpha$

$1 - \alpha = c$

$1 - \alpha = 0.95$

$\alpha = 0.05$

Calculate $z_{\alpha/2}$

$z_{\alpha/2} = z_{0.05/2}$

$z_{\alpha/2} = z_{0.025}$

The z score is:

$z_{\alpha/2} = z_{0.025} =1.96$

The endpoints of the confidence level is:

$(\bar x_1 - \bar x_2) \± z_{\alpha/2} * \sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}$

$2.0 \± 1.96 * \sqrt{\frac{2.1^2}{60}+\frac{3^2}{35}}$

$2.0 \± 1.96* \sqrt{\frac{4.41}{60}+\frac{9}{35}}$

$2.0 \± 1.96 * \sqrt{0.0735+0.2571}$

$2.0 \± 1.96* \sqrt{0.3306}$

$2.0 \± 1.1270$

Split

$(2.0 - 1.1270) \to (2.0 + 1.1270)$

$(0.8730) \to (2.1270)$

Hence, the 95% confidence interval is:

$CI = (0.8730,2.1270)$

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