Question

A recent poll shows that 72% of American adults have some form of social media. A random sample of 19 adults is selected.

a) Find the probability that at least 4 of the adults have social media.

b) Find the probability that at most 7 of the adults have social media.

c) Find the probability that between 6 and 10 of the adults have social media. (including 6 and 10)

d) Find the mean and provide an interpretation.

e) Find the standard deviation.

Answer 1

P(an adult have some form of social media), p = 0.72

q = 1 - p = 0.28

n = 19

P(X < A) = P(Z < (A - mean)/standard deviation)

Mean = np

= 19 x 0.72

= 13.68

Standard deviation =

=

= 1.957

a) P(at least 4) = 1 - P(X < 3.5) (with continuity correction)

= 1 - P(Z < (3.5 - 13.68)/1.957)

= 1 - P(Z < -5.20)

= 1 - 0

= 1

b) P(at most 7) = P(X < 7.5)

= P(Z < (7.5 - 13.68)/1.957)

= P(Z < -3.16)

= 0.0008

c) P(between 6 and 10) = P(X < 10.5) - P(X < 5.5)

= P(Z < (10.5 - 13.68/1.957) - P(Z < (5.5 - 13.68)/1.957)

= P(Z < -1.62) - P(Z < -4.18)

= 0.0526 - 0

= 0.0526

d) Mean = np

= 13.68

In a sample of 19 adults, on average, we expect around 13.68 people to have social media.

e) Standard deviation =

=

= 1.957