adults with smartphones, 48​% use them in meetings. If 15 adult smartphone users are randomly​ selected, find the probability that fewer than 5 of them use their smartphones in meetings

Answer 1

In Excel, we write, = binom.dist(4,15,.48,1) - the 1 indicates cumulative, while 0 is the probability mass function (cumulative gives you the sum of the probabilities from 0 o 4, while probability mass function gives you the value at 4 only; calculators these days can also compute this.

Our answer is the same as that provided above.

Answer 2

We will use the binomial probability.

n choose x = [n!/(x!)*(n-x)!]*p^x(1 - p)^n-x

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

= [15!/(0!)(15!)]*(0.48)⁰(0.52)¹⁵ + [15!/(1!)(14!)]*(0.48)¹(0.52)¹⁴ + [15!/(2!)(13!)]*(0.48)²(0.52)¹³

+ [15!/(3!)(12!)]*(0.48)³(0.52)¹² + [15!/(4!)(11!)]*(0.48)⁴(0.52)¹¹

= 0.000055 + 0.000761 + 0.0049172 + 0.0196687 + 0.0544671 = 0.079869

The probability that fewer than 5 adults use their smartphone during the meetings is about 0.08.

Answer 3

Let x be the number of adults using smart phones in meeting where n=15 is the sample size and p =0.48 is probability of adult using smartphone during meeting

The sample size is not sufficient to utilize the normal approximation to the binomial distribution (np and n(1-p) large, say > 15), so we would have to find the binomial probability for x = 0, 1, 2, 3 and 4, then sum those probabilities to get the total probabilities for x < 5.

Binomial probability for x = k is C(n,k) * p^k * (1-p)^(n-k), where C(n,k) = n!/((n-k)!k!)

So you compute C(15,0) * (0.48)^0 * (0.52)^15 + ... + C(15,4)*(0.48)^4*(0.52)^11 ~= 0.08