a) Let the random variable X= "number of these tracks where SBIRS detects the object." in order to use the binomial probability distribution we need to satisfy some conditions:

1) Independence between the trials (satisfied)

2) A value of n fixed , for this case is 20 (satisfied)

3) Probability of success p =0.2 fixed (Satisfied)

So then we have all the conditions and we can assume that:

$X \sim Bin(n =20, p=0.8)$

b) $X \sim Bin(n =20, p=0.8)$

c) $P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456$

d) $P(X \geq 15) = P(X=15)+ .....+P(X=20)$

$P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456$

$P(X=16)=(20C16)(0.8)^{16} (1-0.8)^{20-16}=0.218$

$P(X=17)=(20C17)(0.8)^{17} (1-0.8)^{20-17}=0.205$

$P(X=18)=(20C18)(0.8)^{18} (1-0.8)^{20-18}=0.137$

$P(X=19)=(20C19)(0.8)^{19} (1-0.8)^{20-19}=0.058$

$P(X=20)=(20C20)(0.8)^{20} (1-0.8)^{20-20}=0.012$

$P(X\geq 15)=0.804208$

e) $E(X) = np = 20*0.8 = 16$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

The probability mass function for the Binomial distribution is given as:

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

Where (nCx) means combinatory and it's given by this formula:

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

Part a

Let the random variable X= "number of these tracks where SBIRS detects the object." in order to use the binomial probability distribution we need to satisfy some conditions:

1) Independence between the trials (satisfied)

2) A value of n fixed , for this case is 20 (satisfied)

3) Probability of success p =0.2 fixed (Satisfied)

So then we have all the conditions and we can assume that:

$X \sim Bin(n =20, p=0.8)$

Part b

$X \sim Bin(n =20, p=0.8)$

Part c

For this case we just need to replace into the mass function and we got:

$P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456$

Part d

For this case we want this probability: $P(X\geq 15)$

And we can solve this using the complement rule:

$P(X \geq 15) = P(X=15)+ .....+P(X=20)$

$P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456$

$P(X=16)=(20C16)(0.8)^{16} (1-0.8)^{20-16}=0.218$

$P(X=17)=(20C17)(0.8)^{17} (1-0.8)^{20-17}=0.205$

$P(X=18)=(20C18)(0.8)^{18} (1-0.8)^{20-18}=0.137$

$P(X=19)=(20C19)(0.8)^{19} (1-0.8)^{20-19}=0.058$

$P(X=20)=(20C20)(0.8)^{20} (1-0.8)^{20-20}=0.012$

$P(X\geq 15)=0.804208$

Part e

The expected value is given by:

$E(X) = np = 20*0.8 = 16$

5 0

3 years ago

What is the equivalent fraction for 0.3333

lorasvet [3.4K]

0.3333 = 1/3

Answer would be: 1/3

6 0

2 years ago

Which is the more reasonable measurement of the distance between the tracks on a railroad: 1.435 ⋅ 10−3 mm or 1.435 ⋅ 103 mm? Ju

SVEN [57.7K]

The best answer is 1.435 ·103 because while they both are really small for a rail road track that one is bigger. the final answer for that is 147.805mm.

5 0

3 years ago

Read 2 more answers

Analyze the diagram below and answer the question that follows. If 2004-02-01-02-00_files/, what is 2004-02-01-02-00_files/? A.

nadya68 [22]

Option A AY || XV given that A, Y, Z are midpoints of sides XW, VW, XV respectively. This can be obtained by knowing what similar triangles are and finding which sides are proportional.

<h3>Find the correct option:</h3>

Similar triangles: If two triangles have proportional sides the are similar.

For example, if ΔABC and ΔDEF are similar then

$\frac{AB}{DE}= \frac{BC}{EF} =\frac{AC}{DF}$

∠ABC = ∠DEF and ∠ACB = ∠DFE

Then we can write that, ΔABC ~ ΔDEF

Here in this question,

Since A, Y, Z are midpoints of sides XW, VW, XV

XA = AW

WY = VY

XZ = VZ

To consider sides AY and XV we should take triangles ΔWAY and ΔWXV

$\frac{WX}{WA} =\frac{2WA}{WA} = 2$ (since A is the midpoint of WX)

$\frac{WV}{WY} =\frac{2WY}{WY} = 2$ (since Y is the midpoint of WV)

∠AWY = ∠XWV (reflexive property)

Therefore ΔWAY and ΔWXV are similar triangles

$\frac{WX}{WA}= \frac{XV}{AY} =\frac{WV}{WY}$ = 2

∠WAY = ∠WXV and ∠AYW = ∠XVW

Hence,

AY || XV option A AY || XV given that A, Y, Z are midpoints of sides XW, VW, XV respectively.

Learn more about similar triangle here:

brainly.com/question/25882965

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Disclaimer: The question was given incomplete on the portal. Here is the complete question.

Question: Analyze the diagram below and answer the question that follows. If Z, Y and A are midpoints of ΔVWX what is true about AY and XY?

A. AY || XV

B. 1/2 AY = XV

C. AY = XV

D. AY ≅ XV

5 0

2 years ago