I would really appreciate it if someone could help me with these last few problems

Ship collisions in the Houston Ship Channel are rare. Suppose the number of collisions are Poisson distributed, with a mean of 1

alexandr1967 [171]

Answer:

$a) \simeq 0.3012$ $b) \simeq 0.0494$ c) $\simeq 0.2438$

Step-by-step explanation:

Rate of collision,

1.2 collisions every 4 months

or, $\frac{1.2}{4}$

= 0.3 collisions per month

So, the Poisson distribution for the random variable no. of collisions per month (X) is given by,

P(X =x) = $\frac{e^{-\lambda}\times {\lambda}^{x}}}{x!}
$

for x ∈ N ∪ {0}

= 0 otherwise --------------------------------------(1)

here, $\lambda = 0.3$ collision / month

No collision over a 4 month period means no collision per month or X =0

Putting X = 0 in (1) we get,

P(X = 0) = $\frac{e^{-0.3}\times {\0.3}^{0}}{0!}
$

$\simeq 0.7408182207$ ------------------------------------(2)

Now, since we are calculating this for 4 months,

so, P(No collision in 4 month period)

= $0.7408182207^{4}$

$\simeq 0.3012$ -----------------------------------------------------------(3)

2 collision in 2 month period means 1 collision per month or X =1

Putting X =1 in (1) we get,

P(X =1) = $\frac{e^{-0.3}\times {\0.3}^{1}}{1!}
$

$\simeq 0.2222454662$ ------------------------------------(4)

Now, since we are calculating this for 2 months, so ,

P(2 collisions in 2 month period)

= $0.2222454662^{2}$

$\simeq 0.0494$ -----------------------------------------(5)

1 collision in 6 months period means

$\frac{1}{6}$ collision per month

Now, P(1 collision in 6 months period)

= P( X = 1/6] (which is to be estimated)

= $\frac {P(X=0)\times 5 + P(X =1)\times 1}{6}$

= \frac {0.7408182207 \times 5 + 0.2222454662 \times 1}{6}[/tex]

$\simeq 0.6543894283$ -------------------------------------------(6)

So,

P(1 collision in 6 month period)

= $0.6543894283^{6}$

$\simeq 0.0785267444$ ------------------------------------------------(7)

So,

P(No collision in 6 months period)

= $(P(X =0)^{6}$

$\simeq 0.1652988882$ ---------------------------------(8)

so,

P(1 or fewer collision in 6 months period)

= (8) + (7 ) = 0.0785267444 +0.1652988882

$\simeq 0.2438$ ---------------------------------------------(9)

7 0

3 years ago

In order to determine the average price of hotel rooms in Atlanta, a sample of 64 hotels was selected. It was determined that th

deff fn [24]

Answer:

The decision rule is

Fail to reject the null hypothesis

The conclusion is

There is no sufficient evidence to show that the average room price is significantly different from $108.50

Step-by-step explanation:

From the question we are told that

The sample size is n = 64

The average price is $\= x = \$ 112$

The population standard deviation is $\sigma = \$ 16$

The level of significance is $\alpha = 0.05$

The population mean is $\mu = \$ 108.5$

The null hypothesis is $H_o : \mu = 108.50$

The alternative hypothesis is $H_a : \mu \ne 108.50$

Generally the test statistics is mathematically represented as

$z = \frac{\= x - \mu }{ \frac{ \sigma}{\sqrt{n} } }$

=> $z = \frac{112 - 108.50 }{ \frac{ 16}{\sqrt{64} } }$

=> $z = 1.75$

From the z table the area under the normal curve to the left corresponding to 1.75 is

$P( Z > 1.75) = 0.040059$

Generally p-value is mathematically represented as

$p-value = 2 * P( Z > 1.75)$

=> $p-value = 2 * 0.040059$

=> $p-value = 0.08012$

From the values obtained we see that $p-value > \alpha$ hence

The decision rule is

Fail to reject the null hypothesis

The conclusion is

There is no sufficient evidence to show that the average room price is significantly different from $108.50

3 0

3 years ago

What is the surface area of s sphere with a radius of 4ft

erik [133]

The formula you use to find the surface area of a sphere is A=4 π r^2
4×π = 12.5663706143
r = 4
4^2 = 16
12.5663706143 × 16 = 201.061929829
Final answer: the surface area is about 201 ft.

8 0

3 years ago

Workout the median 6266 6626 6622 6226 6662 6222 7466

sasho [114]

Answer:

Step-by-step explanation:

the median would be 2,2 just the middle number/numbers if it has 1 # thats the median if it has 2 thats still the median

7 0

3 years ago

Read 2 more answers

What is the largest perimeter possible for a rectangle with positive whole number dimensions and an area of 40 cm?

Vladimir79 [104]

Answer:

C, 82 cm

Step-by-step explanation:

If the area of a rectangle is it's length times it's width and for the problem, it's maximum area must be 40cm than 40x1 would offer the largest perimeter with the smallest area respectively totalling at 82 cm.

8 0

3 years ago