Answer:
[tex]a. \ 0.2107\\\\b.\ 0.6727\\\\c.\ 0.8157[/tex]
Step-by-step explanation:
a. Our random variable follows a binomial distribution which is defined as:
[tex]P(X=x)={n \choose x } \times p^x \times (1-p)^n^-^x[/tex] where [tex]p[/tex] is the probability of success, [tex]n[/tex] the sample size and [tex]x[/tex] the random variable.
-Given that [tex]p=0.32[/tex], [tex]n=10[/tex] what's [tex]P(X=2):[/tex]
[tex]P(X=2)={10\choose 2}\times 0.32^2 \times (1-0.32)^8\\=0.2107[/tex]
Hence, the probability of exactly two is 0.2107
b. The probability that more than two adults are more likely to make purchases during a sales tax holiday is given by:
[tex]P(X>2)=1-P(X=0)-P(X=1)-P(X=2)\\=1-{10\choose 2}\times 0.32^2 \times (1-0.32)^8-{10\choose 1}\times 0.32^1 \times (1-0.32)^9-{10\choose 0}\times 0.32^0 \times (1-0.32)^1^0\\=1-0.2107-0.0995-0.0211\\=0.6727[/tex]
Hence the probability of more that two adult purchases is 0.6727
c. The probability that the number of adults who say they are more likely to make purchases between 2 and 5 is obtain by summing the the probabilities
p(x=2)+p(x=3)+p(x=4)+p(x=5).
-From b above, we already have p(x=2)=0.2107. Therefore:
[tex]P(2\leq X\leq 5)=P(X=2)+P(X=3)+P(X=4)+P(X=5)\\=0.2107+{10\choose 3}\times 0.32^3 \times (1-0.32)^7+{10\choose 4}\times 0.32^4 \times (1-0.32)^6+{10\choose 5}\times 0.32^5 \times (1-0.32)^5\\=0.2107+0.2644+0.2177+0.1229\\=0.8157[/tex]
-Hence the probability of between two and five is 0.8157
