Question

32​% of U.S. adults say they are more likely to make purchases during a sales tax holiday. You randomly select 10 adults. Find the probability that the number of adults who say they are more likely to make purchases during a sales tax holiday is​ (a) exactly​ two, (b) more than​ two, and​ (c) between two and​ five, inclusive

Answer 1

Answer:

$a. \ 0.2107\\\\b.\ 0.6727\\\\c.\ 0.8157$

Step-by-step explanation:

a. Our random variable follows a binomial distribution which is defined as:

$P(X=x)={n \choose x } \times p^x \times (1-p)^n^-^x$ where $p$ is the probability of success, $n$ the sample size and $x$ the random variable.

-Given that $p=0.32$, $n=10$ what's $P(X=2):$

$P(X=2)={10\choose 2}\times 0.32^2 \times (1-0.32)^8\\=0.2107$

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:

$P(X>2)=1-P(X=0)-P(X=1)-P(X=2)\\=1-{10\choose 2}\times 0.32^2 \times (1-0.32)^8-{10\choose 1}\times 0.32^1 \times (1-0.32)^9-{10\choose 0}\times 0.32^0 \times (1-0.32)^1^0\\=1-0.2107-0.0995-0.0211\\=0.6727$

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:

$P(2\leq X\leq 5)=P(X=2)+P(X=3)+P(X=4)+P(X=5)\\=0.2107+{10\choose 3}\times 0.32^3 \times (1-0.32)^7+{10\choose 4}\times 0.32^4 \times (1-0.32)^6+{10\choose 5}\times 0.32^5 \times (1-0.32)^5\\=0.2107+0.2644+0.2177+0.1229\\=0.8157$

-Hence the probability of between two and five is 0.8157