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2 years ago

8

The first term in a sequence is 5. Each new term in the sequence is obtained by adding 4 to the previous term. Determine an equi

valent position-to-term formula. Your answer should be an expression in terms of .

1 answer:

ycow [4]2 years ago

3 0

Answer:

4x+1

Step-by-step explanation:

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Answer:

a) $P(X=2)=(10C2)(0.32)^2 (1-0.32)^{10-2}=0.211$

b) $P(X> 2)=1-P(X\leq 2)=1-[0.0211+0.0995+0.211]=0.668$

c) $P(2 \leq x \leq 5)=0.211+0.264+0.218+0.123=0.816$

Step-by-step explanation:

1) 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".

2) Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=10, p=0.32)$

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

$P(X=2)=(10C2)(0.32)^2 (1-0.32)^{10-2}=0.211$

Part b

$P(X> 2)=1-P(X\leq 2)=1-[P(X=0)+P(X=1)+P(X=2)]$

$P(X=0)=(10C0)(0.32)^0 (1-0.32)^{10-0}=0.0211$

$P(X=1)=(10C1)(0.32)^1 (1-0.32)^{10-1}=0.0995$

$P(X=2)=(10C2)(0.32)^2 (1-0.32)^{10-2}=0.211$

$P(X> 2)=1-P(X\leq 2)=1-[0.0211+0.0995+0.211]=0.668$

Part c

$P(2 \leq x \leq 5)=P(X=2)+P(X=3)+P(X=4)+P(X=5)$

$P(X=2)=(10C2)(0.32)^2 (1-0.32)^{10-2}=0.211$

$P(X=3)=(10C3)(0.32)^3 (1-0.32)^{10-3}=0.264$

$P(X=4)=(10C4)(0.32)^4 (1-0.32)^{10-4}=0.218$

$P(X=5)=(10C5)(0.32)^5 (1-0.32)^{10-5}=0.123$

$P(2 \leq x \leq 5)=0.211+0.264+0.218+0.123=0.816$

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