Question

Assume that when adults with smartphones are randomly selected, 48% use them in meetings or classes. If 14 adult smartphone users are randomly selected, find the probability that fewer than 3 of them use their smartphones in meetings or classes. The probability is . (Type an integer or decimal rounded to four decimal places as needed.)

Answer 1

Solution det X & smartphone users in meetings classes. Here Xo Binomial distribution with parameter (np). The pmf of binomial dist? is, n-> elx-x) = ix20, 1, 2, KPU . 0.0. Here pa 0.98. (1-P) = (-0.49) 0.52

We have to calculate probability that fewer than of 14 adults use their smartphone in meetings of class. 3 i-e. P(x<3) = PlX :o) + P(x+1) +P(X=2) 146.60.98(0.52)'4 + "4470-98)' (0-59)'$ + 140,560.48)”. (0.52) 12 0.00010 + 14. 0.48 0.00020 + 91 0.2304 0.00039 - 0.000• t 0.0014 + 0.00 82 0.0096 je PUX<3): 0.0096 probability is 0.0096

Answer 2

Assume that when adults with smartphones are randomly selected,

37%

use them in meetings or classes. If

15

adult smartphone users are randomly selected, find the probability that exactly

10

of them use their smartphones in meetings or classes.

Answer 3

SOLUTION :

p = adults with smartphones = 48% = 0.48

q = adults without smartphones = 1 -  0.48 = 0.52

As per binomial distribution : For n = 14 :

P(x < 3) 

= P(0) + P(1) + P(2)

= ncr(14, 0)*(0.48)^0 *(0.52)^14 +ncr(14, 1)*(0.48)*(0.52)^13

+ncr(14, 2)*(0.48)^2 * (0.52)^12

= 0.0097 (ANSWER).