Question

53​% of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults. Find the probability that the number of U.S. adults who have very little confidence in newspapers is ​ (a) exactly​ five, (b) at least​ six, and​ (c) less than four. ​(Round to three decimal places as​ needed.)

Answer 1

Answer:

a

   $P(X = 5) = 0.2423$

b

    $P( X \ge 6 )= 0.4516$

c

   $P( X < 4 )= 0.12694$

Step-by-step explanation:

From the question we are told that

   The population proportion is   p = 0.53

    The sample size is  n = 10

Generally the distribution of the confidence of US adults in newspapers follows a binomial distribution  

i.e  

         $X \~ \ \ \ B(n , p)$

and the probability distribution function for binomial  distribution is  

      $P(X = x) = ^{n}C_x * p^x * (1- p)^{n-x}$

Here C stands for combination hence we are going to be making use of the combination function in our calculators        

Generally the probability that the number of U.S. adults who have very little confidence in newspapers is exactly​ five is mathematically represented as

     $P(X = 5) = ^{10}C_5 * 0.53^5 * (1- 0.53)^{10-5}$

=>  $P(X = 5) = 252 * 0.04182 * 0.023$

=>  $P(X = 5) = 0.2423$

Generally the probability that the number of U.S. adults who have very little confidence in newspapers is  at least​ six is mathematically represented as

  $P( X \ge 6 )= P( X = 6) + P( X = 7) + P( X = 8) + P( X = 9) + P( X = 10)$

=> $P( X \ge 6 )= [^{10}C_6 * [0.53]^6 * (1- 0.53)^{10-6}] + [^{10}C_7 * [0.53]^7 * (1- 0.53)^{10-7}] + [^{10}C_8 * [0.53]^8 * (1- 0.53)^{10-8}] + [^{10}C_9 * [0.53]^9 * (1- 0.53)^{10-9}] + [^{10}C_{10} * [0.53]^{10} * (1- 0.53)^{10-10}]$

=> $P( X \ge 6 )= [0.227] + [0.1464] + [0.0619] + [0.0155] + [0.00082]$

=> $P( X \ge 6 )= 0.4516$

Generally the probability that the number of U.S. adults who have very little confidence in newspapers is  less than four is mathematically represented as

    $P( X < 4 )= P( X = 3) + P( X = 2) + P( X = 1) + P( X = 0)$

=>  $P( X < 4 )= [^{10}C_3 * 0.53^3 * (1- 0.53)^{10-3}] + [^{10}C_2 * 0.53^2 * (1- 0.53)^{10-2}] + [^{10}C_1 * 0.53^1 * (1- 0.53)^{10-1}] + [^{10}C_0 * 0.53^0 * (1- 0.53)^{10-0}]$

=>  $P( X < 4 )= 0.09051 + 0.03 + 0.0059 + 0.000526$

=>  $P( X < 4 )= 0.12694$