Question 1194301
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n = 10 = sample size
p = 0.70 = probability of germination


We have a binomial distribution problem for the following reasoning<ul><li>There are two outcomes: Either a seed germinates, or it does not.</li><li>Each trial (seed) is independent of one another. No one seed affects the germination of another.</li><li>The probability of success (ie germination) is the same for each trial, which is p = 0.70</li><li>There are a fixed number of trials (n = 10).</li></ul>We'll use the aptly named binomial probability formula
P(x) = (n C x)*(p)^x*(1-p)^(n-x)
the values of n and p were mentioned earlier
The n C x refers to the nCr formula


The x values take on items from the set {0,1,2,3,4,5,6,7,8,9,10}
because we'll define x like so
x = number of seeds that germinate


Let's compute P(x) when x = 0
P(x) = (n C x)*(p)^x*(1-p)^(n-x)
P(0) = (10 C 0)*(0.70)^0*(1-0.70)^(10-0)
P(0) = (1)*(0.70)^0*(1-0.70)^(10-0)
P(0) = 0.0000059049


Do the same for x = 1
Keep n = 10 and p = 0.7 the same
P(x) = (n C x)*(p)^x*(1-p)^(n-x)
P(1) = (10 C 1)*(0.70)^1*(1-0.70)^(10-1)
P(1) = (10)*(0.70)^1*(1-0.70)^(10-1)
P(1) = 0.000137781


This process is repeated for the other x values from x = 2 all the way up to x = 10.
Spreadsheet software makes quick work of the computations, and it gives a natural easy way to display it as a table.<table border = "1" cellpadding = "5"><tr><td>x</td><td>P(x)</td></tr><tr><td>0</td><td>0.0000059049</td></tr><tr><td>1</td><td>0.000137781</td></tr><tr><td>2</td><td>0.0014467005</td></tr><tr><td>3</td><td>0.009001692</td></tr><tr><td>4</td><td>0.036756909</td></tr><tr><td>5</td><td>0.1029193452</td></tr><tr><td>6</td><td>0.200120949</td></tr><tr><td>7</td><td>0.266827932</td></tr><tr><td>8</td><td>0.2334744405</td></tr><tr><td>9</td><td>0.121060821</td></tr><tr><td>10</td><td>0.0282475249</td></tr></table>
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The previous section is a lot of set up if you aren't quite familiar with the binomial distribution. 
Fortunately, software can be used to quickly generate such a table.


We'll use that table to answer parts (a) through (f)


For part (a), the phrasing "at most" means "that is the highest we can go". It is the ceiling value.
"At most 7" means "7 is the highest we can go".


We're tasked to find P(x ≤ 7)
This is the same as adding P(0) all the way through to P(7)
That's 8 numbers we have to add up.


But we can take a shortcut. Notice that
P(x ≥ 8) = P(8) + P(9) + P(10)
P(x ≥ 8) = 0.2334744405 + 0.121060821 + 0.0282475249
P(x ≥ 8) = 0.3827827864


Then we can say,
P(x ≤ 7) + P(x ≥ 8) = 1
P(x ≤ 7) = 1 - P(x ≥ 8)
P(x ≤ 7) = 1 - 0.3827827864
P(x ≤ 7) = 0.6172172136



Answer: Approximately 0.6172172136
Round this however you need to


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Part (b)


"At least 4" means "4 or more".
We want to compute P(x ≥ 4)


Like before we can use a shortcut to find P(x ≤ 3)
P(x ≤ 3) = P(0) + P(1) + P(2) + P(3)
P(x ≤ 3) = 0.0000059049 + 0.000137781 + 0.0014467005 + 0.009001692
P(x ≤ 3) = 0.0105920784


Then,
P(x ≤ 3) + P(x ≥ 4) = 1
P(x ≥ 4) = 1 - P(x ≤ 3)
P(x ≥ 4) = 1 - 0.0105920784
P(x ≥ 4) = 0.9894079216
The longer alternative method would be to add up P(4) all the way up to P(10) which means you have to add up 7 values. 


Answer: Approximately 0.9894079216


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Part (c)


P(3 ≤ x ≤ 7) = P(3) + P(4) + P(5) + P(6) + P(7)
P(3 ≤ x ≤ 7) = 0.009001692 + 0.036756909 + 0.1029193452 + 0.200120949 + 0.266827932
P(3 ≤ x ≤ 7) = 0.6156268272


Answer: Approximately 0.6156268272


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Part (d)


P(4 < x < 9) = P(5 ≤ x ≤ 8)
P(4 < x < 9) = P(5) + P(6) + P(7) + P(8)
P(4 < x < 9) = 0.1029193452 + 0.200120949 + 0.266827932 + 0.2334744405
P(4 < x < 9) = 0.8033426667


Answer: Approximately 0.8033426667


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I'll let you tackle parts (e) and (f)


Hints:
P(x < 5) = P(0) + P(1) + P(2) + P(3) + P(4)
P(x > 3) = P(4) + P(5) + P(6) + P(7) + P(8) + P(9) + P(10)
Or you can take the shortcut that
P(x ≤ 3) + P(x > 3) = 1
where
P(x ≤ 3) = P(0) + P(1) + P(2) + P(3)
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