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SVETLANKA909090 [29]
3 years ago
12

Suppose that x has a binomial distribution with n = 201 and p = 0.45. (Round np and n(1-p) answers to 2 decimal places. Round yo

ur answers to 4 decimal places. Round z values to 2 decimal places. Round the intermediate value (σ) to 4 decimal places.) (a) Show that the normal approximation to the binomial can appropriately be used to calculate probabilities about x.
Mathematics
1 answer:
ycow [4]3 years ago
3 0

Answer:

a) It can be used because np and n(1-p) are both greater than 5.

Step-by-step explanation:

Binomial distribution and approximation to the normal:

The binomial distribution has two parameters:

n, which is the number of trials.

p, which is the probability of a success on a single trial.

If np and n(1-p) are both greater than 5, the normal approximation to the binomial can appropriately be used.

In this question:

n = 201, p = 0.45

So, lets verify the conditions:

np = 201*0.45 = 90.45 > 5

n(1-p) = 201*(1-0.45) = 201*0.55 = 110.55 > 5

Since both np and n(1-p) are greater than 5, the approximation can be used.

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3 years ago
Use Pascal’s Triangle to expand the binomial. (8v+s)^5
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<span>Pascal 's triangle is 
</span>

1<span>
</span><span><span>1,1</span><span>
</span></span><span><span>1,2,1</span><span>
</span></span><span><span>1,3,3,1</span><span>
</span></span><span><span>1,4,6,4,1</span><span>
</span></span><span>1,5,10,10,5,1</span>

the row <span>1,5,10,10,5,1</span> is the one we need to expand <span><span><span>(8</span>v+s)</span></span>⁵.

Given <span><span>(a+b)</span></span>⁵ each term of the row is the coefficient of akbt

with k goes form 5 to 0 and t goes from 0 to 5.

so <span><span><span><span>(a+b)</span></span></span></span>⁵=1a⁵b⁰+5a⁴b¹+10a³b²+10a²b³+5a¹b⁵+1a⁰b⁵

<span><span>=<span>a</span></span></span>⁵+5a⁴b+10a³b²+10a²b³+5ab⁵+b⁵.

In the case of <span><span><span>(8</span>v+s)</span></span>⁵:

<span><span>a<span>=8</span>v</span><span>
</span></span><span>b=s</span>

<span><span>1<span><span>(8</span>v)</span></span></span>⁵s⁰+5(8v)⁴s¹+10(8v)³s²+10(8v)²s³+5(8v)¹s⁵+1(8v)⁰s⁵

<span> <span><span><span>=<span><span> (8</span>v)</span></span></span></span>⁵+5(8v)⁴s+10(8v)³s²+10(8v)²s³+5(8v)s⁵+s⁵=</span>

<span><span><span>= 8</span></span></span>⁵v⁵+5⋅8⁴v⁴s+10⋅8³v³s²+10⋅8²v²s³+40vs⁵+s⁵=

<span><span><span>= 32768</span><span><span>v</span></span></span></span>⁵<span><span><span>+20480</span><span><span>v</span></span></span></span>⁴<span><span><span>s</span><span>+5120</span><span><span>v</span></span></span></span>³<span><span><span><span>s</span></span></span></span>²<span><span><span>+640</span><span><span>v</span></span></span></span>²<span><span><span><span>s</span></span></span></span>³<span><span><span>+40</span><span>v<span>s</span></span></span></span>⁵<span><span><span>+</span><span>s</span></span></span>⁵<span>.</span>
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