Question

In: Statistics and Probability

Suppose a random sample of 100 households were selected from a large city. if the true...

Suppose a random sample of 100 households were selected from a large city. if the true proportion of households in the city that owns at least one pet is 0.4, then what is the probability that at least 50% of the sample owns at least one pet?

a) 0.0207

b) 0.9793

c) 0.5793

d) 0.4207

e) 0.2578

Expert Solution

Solution:

Given that,

n = 100

= 0.40

1 - = 1 - 0.40 = 0.60

So,

a ) = = 0.55

= $\sqrt$ ( 1 - ) / n

= $\sqrt$ 0.40 * 0.60 / 100

= 0.0490

p ( > 0.50 )

= 1 - p ( < 0.50 )

= 1 - p ( - / ) < ( 0.50 - 0.40 / 0.0490 )

= 1 - p ( z < 0.10 / 0.0490 )

= 1 - p ( z < 2.04 )

Using z table

= 1 - 0.9793

= 0.0207

Probability = 0.0207

option a ) is correct.

orchestra answered 2 years ago

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