Question

According to a study conducted by a statistical​ organization, the proportion of people who are satisfied with the way things are going in their lives is 0.84. Suppose that a random sample of 100 people is obtained. Complete parts​ (a) and​ (b) below.

​(a)  What is the probability that the proportion who are satisfied with the way things are going in their life exceeds 0.88​?

The probability that the proportion who are satisfied with the way things are going in their life exceeds 0.88 is ___.

​(Round to four decimal places as​ needed.)

​(b) Would it be unusual for a survey of 100 people to reveal that 77 or fewer people in the sample are satisfied with their​ lives?

The probability that 77 or fewer people in the sample are satisfied is ___​, which_ _(is not/is)__unusual because this probability __(is not/is)__ less than ___(5/ 0.05/ 50/ 0.5)___

​(Round to four decimal places as​ needed.)

Answer 1

a)

P( < p) = P(Z < ( - p) / sqrt [ p ( 1 - p) / n ]

So,

P( > 0.88) = P(Z > (0.88 - 0.84) / sqrt [ 0.84 ( 1 - 0.84) / 100 ] )

= P(Z > 1.09)

= 0.1379

b)

Mean = np = 100 * 0.84 = 84

Standard deviation = sqrt [ n p ( 1 - p) ] = sqrt [ 100 * 0.84 ( 1 - 0.84) ] = 3.6661

Using normal approximation,

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

With continuity correction,

P(X <= 77) = P(X < 77.5)

P ( ( X < 77.5 ) = P ( Z < 77.5 - 84 ) / 3.6661 )
= P ( Z < -1.77 )
P ( X < 77.5 ) = 0.0384

The probability that 77 or fewer people in the sample are satisfied is 0.0384 is unusual

because this probability is less than 0.05