Question

10. Suppose 29% of children prefer strawberry ice cream to chocolate. We randomly pick a sample of 1,000 children.

1 What is the probability that between 24% and 32% of the children in our sample prefer strawberry ice cream? Blank 1

2 What is the probability that less than 27% of the children in our sample prefer strawberry ice cream? Blank 2

Answer #1

1)

Using normal approximation,

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

So,

P(0.24 < < 0.32) = P( < 0.32) - P( < 0.24)

= P(Z < ( 0.32 - 0.29) / sqrt ( 0.29 ( 1 - 0.29) / 1000) ) - P(Z < ( 0.24 - 0.29) / sqrt ( 0.29 ( 1 - 0.29) / 1000) )

= P(Z < 2.09) - P(Z < -3.48)

= 0.9817 - 0.0003 (From Z table)

= **0.9814**

b)

P( < 0.27) = P(Z < ( 0.27 - 0.29) / sqrt ( 0.29 ( 1 - 0.29) / 1000) )

= P(Z < -1.39)

= **0.0823** (From Z table)

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