Question

use binomial tables
the local school district is suggesting that the school year increases to 11 months a year instead of nine months. sixty - five percent of citizens support the change. eight citizens are surveyed. use four decimal places when reading table.
what is the probability that no more than 1 in a random sample of eight citizens support increasing the school year? select
what is the probability that more than 7 in a random sample of eight citizens support the school year? select
what is the expected value? select
what is p(2<=x<=4)? select

Explanation:

Step1: Identify binomial parameters

Given \(n = 8\) (number of trials, i.e., citizens surveyed), \(p=0.65\) (probability of success, i.e., a citizen supports the change).

Step2: Probability no more than 1 (\(P(X\leq1)\))

Using binomial formula \(P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}\), where \(C(n,k)=\frac{n!}{k!(n - k)!}\)
\(P(X\leq1)=P(X = 0)+P(X = 1)\)
\(P(X = 0)=C(8,0)\times(0.65)^{0}\times(0.35)^{8}=\frac{8!}{0!(8 - 0)!}\times1\times0.35^{8}=1\times1\times0.00022518=0.0002\)
\(P(X = 1)=C(8,1)\times(0.65)^{1}\times(0.35)^{7}=\frac{8!}{1!(8 - 1)!}\times0.65\times0.35^{7}=8\times0.65\times0.00064337=0.0033\)
\(P(X\leq1)=0.0002 + 0.0033=0.0035\)

Step3: Probability more than 7 (\(P(X>7)\))

\(P(X>7)=P(X = 8)\)
\(P(X = 8)=C(8,8)\times(0.65)^{8}\times(0.35)^{0}=\frac{8!}{8!(8 - 8)!}\times0.65^{8}\times1=1\times0.03198004=0.0320\)

Step4: Expected value (\(E(X)\))

For binomial distribution \(E(X)=n\times p\), so \(E(X)=8\times0.65 = 5.2000\)

Step5: \(P(2\leq X\leq4)\)

\(P(2\leq X\leq4)=P(X = 2)+P(X = 3)+P(X = 4)\)
\(P(X = 2)=C(8,2)\times(0.65)^{2}\times(0.35)^{6}=\frac{8!}{2!(8 - 2)!}\times0.4225\times0.00183826=28\times0.4225\times0.00183826=0.0218\)
\(P(X = 3)=C(8,3)\times(0.65)^{3}\times(0.35)^{5}=\frac{8!}{3!(8 - 3)!}\times0.274625\times0.00521539=56\times0.274625\times0.00521539=0.0808\)
\(P(X = 4)=C(8,4)\times(0.65)^{4}\times(0.35)^{4}=\frac{8!}{4!(8 - 4)!}\times0.17850625\times0.01500625=70\times0.17850625\times0.01500625=0.1875\)
\(P(2\leq X\leq4)=0.0218+0.0808 + 0.1875=0.2901\)

Answer:

• Probability no more than 1: \(0.0035\)
• Probability more than 7: \(0.0320\)
• Expected value: \(5.2000\)
• \(P(2\leq X\leq4)\): \(0.2901\)