Question

The National Coffee Association reported that 63% of U.S. adults drink coffee daily. A large hotel chain randomly selects 250 customers (assume U.S. adults) for a survey. One of the questions asks if the person taking the survey drank coffee daily. Would it be unusual if less than 142 of the sampled adults drank coffee daily? What is the random variable? Is it plausible to assume that the distribution of the random variable is approximately normal? Explain. If appropriate, determine the distribution of the random variable and compute the indicated probability. Would it be unusual if less than 142 of the sampled adults drank coffee daily? Explain.

Answer 1

X ~ Bin ( n , p)

Where n = 250 , p = 0.63

n(1-p) = 250 * 0.37 = 92.3

Mean = n p = 250 * 0.63 = 157.5

Standard deviation = sqrt [ n p ( 1 - p) ]

= sqrt [ 250 * 0.63 ( 1 - 0.63) ]

= 7.6338

Since np > 5 and n ( 1-p) > 5 , normal approximation is appropriate.

Using normal approximation,

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

With continuity correction,

P(X < 142) = P(Z < (141.5 - 157.5) / 7.6338 )

= P(Z < -2.10)

= 0.0179 (From Z table)

Since this probability is less than 0.05, the event is unusual.