 Alfred Martin

2021-12-16

According to a renowned expert, heavy smokers make up 70% of lung cancer patients. If his claim is true, calculate the likelihood

(a) That 10 of these people were recently admitted to

a hospital.
A majority smoke heavily (more than half).
(b) There are exactly 4 heavy smokers.
(c) Less than two people do not smoke. Linda Birchfield

Expert

(a) Find the likelihood that more than half of 10 of these recently hospitalized patients are heavy smokers.
The following calculation shows the likelihood that, out of 10 such patients who were recently admitted to a hospital, more than half are heavy smokers:
Let X represent the proportion of recently admitted patients who smoke heavily and who have a binomial distribution with a success rate of 0.70 and a sample size of 10 patients.
That is, $n=10,p=0.70,q=0.30\left(=1-0.30\right)$
It follows that, for the probability distribution,

where n is the number of tries and p is the likelihood that each experiment will be successful.
$P\left(X>5\right)=1-P\left(X\le 5\right)$
To determine the chance that x = 5, use Excel.
To get the P-value, follow the instructions:
1. Open EXCEL
2. Go to Formula bar.
3. In formula bar enter the function as“=BINOMDIST”
4. Enter the number of success as 5.
5. Enter the Trails as 10.
6. Enter the probability as 0.70.
7. Enter the cumulative as True.
8. Click enter.
EXCEL output:
The Excel output indicates that the P-value is 0.1503.
$P\left(X>5\right)=1-P\left(X\le 5\right)$
$=1-0.1503$
$=0.8497$
There is a 0.8497 percent chance that out of 10 of these individuals who were recently admitted to a hospital, more than half are habitual smokers.
(b) Find the likelihood that, among 10 such patients who were just admitted to a hospital, 4 are heavy smokers.
The following calculation shows the likelihood that, out of 10 such patients who were just admitted to a hospital, exactly 4 are heavy smokers:
To determine the likelihood that x equals 4, use Excel.
To get the P-value, follow the instructions:
1. Open EXCEL
2. Go to Formula bar.
3. In formula bar enter the function as“=BINOMDIST”
4. Enter the number of success as 4.
5. Enter the Trails as 10.
6. Enter the probability as 0.70.
7. Enter the cumulative as False
8. Click enter.
EXCEL output:
The P-value, as shown by the Excel result, is 0.0368.
There is a 0.0368 percent chance that out of 10 recently hospitalized patients, 4 are heavy smokers.
(c) Find the possibility that less than two of every 10 of these recently hospitalized patients who are not smokers.
The following calculation shows the likelihood that, out of 10 patients recently admitted to a hospital, fewer than 2 are non-smokers:
Excel may be used to calculate the likelihood that x = 1.
To get the P-value, follow the instructions:
1. Open EXCEL
2. Go to Formula bar.
3. In formula bar enter the function as“=BINOMDIST”
4. Enter the number of success as 1.
5. Enter the Trails as 10.
6. Enter the probability as 0.30.
7. Enter the cumulative as True.
8. Click enter.
EXCEL output:
The P-value obtained from the Excel output is 0.1493.
There is a 0.1493 percent chance that out of 10 of these individuals who were recently admitted to a hospital, fewer than two are nonsmokers. Charles Benedict

Expert

Step 1
Given that :
Let X be the number of patients who are heavy smokers.
$C\sim \text{Binomial(10, 0.70}\right)$
Step 2
a) $P\left(x>5\right)=\sum _{x=6}^{10}\left(\begin{array}{c}10\\ x\end{array}\right)\left(0.70{\right)}^{x}\left(0.30{\right)}^{10-x}$
$=\left(\begin{array}{c}10\\ 6\end{array}\right)\left(0.70{\right)}^{6}\left(0.30{\right)}^{4}+\left(\begin{array}{c}10\\ 7\end{array}\right)\left(0.70{\right)}^{7}\left(0.30{\right)}^{3}$
$+\left(\begin{array}{c}10\\ 8\end{array}\right)\left(0.70{\right)}^{8}\left(0.30{\right)}^{2}+\left(\begin{array}{c}10\\ 9\end{array}\right)\left(0.70{\right)}^{9}\left(0.30{\right)}^{1}+\left(\begin{array}{c}10\\ 10\end{array}\right)\left(0.70{\right)}^{10}\left(0.30{\right)}^{0}$
$=0.20012+0.26683+0.23347+0.12106+0.02825$
$=0.8497$
Step 2
b) $P\left(X=4\right)=\left(\begin{array}{c}10\\ 4\end{array}\right)\left(0.70{\right)}^{4}\left(0.30{\right)}^{6}$
$=0.0368$
c) $P\left(X<2\right)=P\left(x=0\right)+P\left(X=1\right)$
$=\left(\begin{array}{c}10\\ 0\end{array}\right)\left(0.70{\right)}^{0}\left(0.30{\right)}^{10}+\left(\begin{array}{c}10\\ 1\end{array}\right)\left(0.70{\right)}^{1}\left(0.30{\right)}^{9}$
$=0.0000059+0.0001378$
$=0.0001437$ nick1337

Expert

Solution: 5.7p=0.7
a) For
b) For

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