Question

According to a recent report, 68% of Internet searches in a particular month used the Google search engine. Assume that a sample of 21 searches is studied. Round the answers to four decimal places.

(a) What is the probability that exactly 17 of them used Google?

(b) What is the probability that 12 or fewer used Google?

(c) What is the probability that more than 17 of them used Google?

(d) Would it be unusual if fewer than 9 used Google?

It ▼(Choose one) be unusual if fewer than 9used Google since the probability is _. .

Answer 1

X ~ Bin ( n , p )

Where n = 21 , p = 0.68

Mean = n p = 21 * 0.68 = 14.28

Standard deviation = sqrt [ n p ( 1 - p) ] = sqrt [ 21 * 0.68 ( 1 - 0.68) ) = 2.1377

Using normal approximation

P(X < x) = P( Z < ( X - mean ) / SD )

With continuity correction

P(X = 17) = P(16.5 < X < 17.5)
P ( 16.5 < X < 17.5 ) = P ( Z < ( 17.5 - 14.28 ) / 2.1377 ) - P ( Z < ( 16.5 - 14.28 ) / 2.1377 )
= P ( Z < 1.51) - P ( Z < 1.04 )
= 0.9345 - 0.8508
= 0.0836

b)

Using normal approximation with continuity correction

P(X <= 12) = P(X < 12.5)

P ( ( X < 12.5 ) = P ( Z < 12.5 - 14.28 ) / 2.1377 )
= P ( Z < -0.83 )
P ( X < 12.5 ) = 0.2033

c)

Using normal approximation with continuity correction

P(X >= 17) = P(X > 16.5)

P ( X > 16.5 ) = P(Z > (16.5 - 14.28 ) / 2.1377 )
= P ( Z > 1.04 )
= 1 - P ( Z < 1.04 )
= 1 - 0.8508
= 0.1492

d)

Using normal approximation with continuity correction

P(X < 9) = P(X < 8.5)

P ( ( X < 8.5 ) = P ( Z < 8.5 - 14.28 ) / 2.1377 )
= P ( Z < -2.7 )
P ( X < 8.5 ) = 0.0035

Since this probability less than 0.05, it is unusual .

It would be unusual if fewer than 9 used Google since the probability is less than 0.05