Question

find the probability, to 4 decimal places.
it is possible when rounded that a probability is 0.0000
d) exactly none are left - handed.

e) exactly 10 are left - handed.

f) at least 4 are left - handed.

g) at most 3 are left - handed.

h) at least 6 are left - handed.

i) is 6 an unusually high number of people that are left - handed in a s
no, because ( p ( x = 6 ) > 0.05 )
question help:

Explanation:

To solve these probability problems, we assume this is a binomial probability problem (since we're dealing with the number of left - handed people, which is a binary outcome: left - handed or not left - handed). The binomial probability formula is \(P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}\), where \(C(n,k)=\frac{n!}{k!(n - k)!}\), \(n\) is the number of trials, \(k\) is the number of successful trials, and \(p\) is the probability of success on a single trial. However, since the problem does not provide the values of \(n\) (the number of people in the sample) and \(p\) (the probability that a person is left - handed), we can't calculate the exact probabilities. But we can provide the general steps:

Step 1: Identify \(n\), \(p\), and \(k\)

We need to know the total number of people \(n\) and the probability \(p\) that a single person is left - handed. For example, if we assume that the proportion of left - handed people in the population \(p = 0.1\) (a common estimate) and \(n\) is given (say \(n = 10\) for simplicity, although this is an assumption), and we want to find \(P(X=0)\) (exactly none are left - handed):

Step 2: Calculate the combination \(C(n,k)\)

For \(k = 0\) and \(n\) (let's say \(n = 10\)), \(C(10,0)=\frac{10!}{0!(10 - 0)!}=\frac{10!}{1\times10!}=1\)

Step 3: Calculate \(p^{k}\) and \((1 - p)^{n - k}\)

If \(p = 0.1\) and \(k = 0\), \(p^{k}=(0.1)^{0}=1\) and \((1 - p)^{n - k}=(1 - 0.1)^{10-0}=(0.9)^{10}\approx0.3487\)

Step 4: Calculate \(P(X = k)\)

\(P(X = 0)=C(10,0)\times(0.1)^{0}\times(0.9)^{10}=1\times1\times0.3487 = 0.3487\) (this is just an example with assumed \(n\) and \(p\))

For the other parts:

• For \(P(X = 10)\) (exactly 10 are left - handed), \(k = 10\), \(C(n,10)=\frac{n!}{10!(n - 10)!}\), \(p^{10}\) and \((1 - p)^{n - 10}\)
• For \(P(X\geq4)\) (at least 4 are left - handed), we calculate \(P(X = 4)+P(X = 5)+\cdots+P(X = n)\)
• For \(P(X\leq3)\) (at most 3 are left - handed), we calculate \(P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3)\)
• For \(P(X\geq6)\) (at least 6 are left - handed), we calculate \(P(X = 6)+P(X = 7)+\cdots+P(X = n)\)
• For the question about whether 6 is an unusually high number, we use the rule that a value \(k\) is unusually high if \(P(X\geq k)<0.05\)

Since the problem does not provide the necessary values of \(n\) (sample size) and \(p\) (probability of a person being left - handed), we can't give the final numerical answers. If you provide the values of \(n\) and \(p\), we can calculate the probabilities using the binomial probability formula as shown above.

Answer:

To solve these probability problems, we assume this is a binomial probability problem (since we're dealing with the number of left - handed people, which is a binary outcome: left - handed or not left - handed). The binomial probability formula is \(P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}\), where \(C(n,k)=\frac{n!}{k!(n - k)!}\), \(n\) is the number of trials, \(k\) is the number of successful trials, and \(p\) is the probability of success on a single trial. However, since the problem does not provide the values of \(n\) (the number of people in the sample) and \(p\) (the probability that a person is left - handed), we can't calculate the exact probabilities. But we can provide the general steps:

Step 1: Identify \(n\), \(p\), and \(k\)

We need to know the total number of people \(n\) and the probability \(p\) that a single person is left - handed. For example, if we assume that the proportion of left - handed people in the population \(p = 0.1\) (a common estimate) and \(n\) is given (say \(n = 10\) for simplicity, although this is an assumption), and we want to find \(P(X=0)\) (exactly none are left - handed):

Step 2: Calculate the combination \(C(n,k)\)

For \(k = 0\) and \(n\) (let's say \(n = 10\)), \(C(10,0)=\frac{10!}{0!(10 - 0)!}=\frac{10!}{1\times10!}=1\)

Step 3: Calculate \(p^{k}\) and \((1 - p)^{n - k}\)

If \(p = 0.1\) and \(k = 0\), \(p^{k}=(0.1)^{0}=1\) and \((1 - p)^{n - k}=(1 - 0.1)^{10-0}=(0.9)^{10}\approx0.3487\)

Step 4: Calculate \(P(X = k)\)

\(P(X = 0)=C(10,0)\times(0.1)^{0}\times(0.9)^{10}=1\times1\times0.3487 = 0.3487\) (this is just an example with assumed \(n\) and \(p\))

For the other parts:

• For \(P(X = 10)\) (exactly 10 are left - handed), \(k = 10\), \(C(n,10)=\frac{n!}{10!(n - 10)!}\), \(p^{10}\) and \((1 - p)^{n - 10}\)
• For \(P(X\geq4)\) (at least 4 are left - handed), we calculate \(P(X = 4)+P(X = 5)+\cdots+P(X = n)\)
• For \(P(X\leq3)\) (at most 3 are left - handed), we calculate \(P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3)\)
• For \(P(X\geq6)\) (at least 6 are left - handed), we calculate \(P(X = 6)+P(X = 7)+\cdots+P(X = n)\)
• For the question about whether 6 is an unusually high number, we use the rule that a value \(k\) is unusually high if \(P(X\geq k)<0.05\)

Since the problem does not provide the necessary values of \(n\) (sample size) and \(p\) (probability of a person being left - handed), we can't give the final numerical answers. If you provide the values of \(n\) and \(p\), we can calculate the probabilities using the binomial probability formula as shown above.