Question

According to a recent study, 40% of children under the age of 2 have used a mobile device as a form of entertainment (i.e play video games, watch movies, listen to, music, etc.) If a random sample of 200 children under the age of 2 is selected determine the probability that: a) at least 45 % have used a mobile device as a form of entertainment. b) between 37% and 48% have used a mobile device as a form of entertainment c) at least 55% have not used a mobile device as a form of entertainment

Answer 1

Solution

Back-up Theory

If X ~ B(n, p). i.e., X has Binomial Distribution with parameters n and p, where n = number of trials and

p = probability of one success, then, probability mass function (pmf) of X is given by

p(x) = P(X = x) = (ⁿC_x)(p^x)(1 - p)^{n – x}, x = 0, 1, 2, ……. , n ………….........................................................................………..(1)

[This probability can also be directly obtained using Excel Function: Statistical, BINOMDIST....................………...........….(1a)

Mean (average) of X = E(X) = µ = np….....................................................................……………………………………………..(2)

Variance of X = V(X) = σ² = np(1 – p)………….................................................................………………………………………..(3)

Standard Deviation of X = SD(X) = σ = √{np(1 – p)} ……......................................................…………………………….……...(4)

If X ~ B(n, p), np ≥ 10 and np(1 - p) ≥ 10, then Binomial probability can be approximated by

Standard Normal probabilities by Z = (X – np)/√{np(1 - p)} ~ N(0, 1) …………………,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,.......,,…..(5)

Probability values for the Standard Normal Variable, Z, can be directly read off from Standard Normal Tables................... (6a)

or can be found using Excel Function: Statistical, NORMSDIST which gives P(Z ≤ z) …..............................................……(6b)

Now, to work out the solution,

Let X = Number of children out of a random sample of 200 children under the age of 2 who have

used a mobile device as a form of entertainment. Then, X ~ B(200, 0.4) ............................................................................. (7)

[0.4 comes from ‘40% of children under the age of 2 have used a mobile device as a form of entertainment’]

Since both np = 80 ≥ 10 and np(1 - p) = 48 ≥ 10, Binomial probability can be approximated by

Standard Normal probabilities, vide (5) ................................................................................................................................. (8)

Part (a)

45% of 200 = 90.

Probability that at least 45% have used a mobile device as a form of entertainment

= P(X ≥ 90)

= P[Z ≥ {(90 – 80)/√48}] [vide (8) and (5)]

= P(Z ≥ 1.4434)

= 0.0745 [vide (6b)] Answer 1

Part (b)

37% of 200 = 74 and 48% of 200 = 96.

Probability that between 37% and 48% have used a mobile device as a form of entertainment

= P(74 ≤ X ≤ 96)

= P[{(74 – 80)/√48} ≤ Z ≤ {(96 – 80)/√48}] [vide (8) and (5)]

= P[Z ≤ {(96 – 80)/√48}] – P[Z {(74 – 80)/√48}]

= P(Z ≤ 2.3094) - P(Z ≤ -0.8661)

= 0.9895 – 0.1948 [vide (6b)]

= 0.7947 Answer 2

Part (c)

Probability that at least 55% have not used a mobile device as a form of entertainment

= Probability that at most 45% have  used a mobile device as a form of entertainment

= P(X ≤ 90)

= (1 – 0.0745) [from Answer 1]

= 0.9225 Answer 3

DONE