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24% of all students enjoy the school lunch. If there are 2000 students at the school,...

24% of all students enjoy the school lunch. If there are 2000 students at the school, what is the approximate probability that at least 23% enjoyed the school lunch?

Question 1 options:

.1475

0.018

0.50

0.982

.8525
math   Statistics-And-Probability   
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Answer #1

Answer

population proportion Po 0 = 24/100 = 0.24

sample proportion hat{p} = 23/100 = 0.23

sample size is n = 2000

we have to find the probability of at least 0.23 proportion means probability of greater than 0.22

P(p > 0.22) = P(z > (p-po)/ V (po * (1-Po))/n)

setting the given values, we get

P(p>0.22)- P(z >(0.22 0.24)/(0.24(1-0.24))/2000)

this gives us

P(p > 0.22)-P(z > (-0.02)70.00955)-P(z > _2.094)

Using identity P(z>-a) = P(z<a)

we can write it as

P(p > 0.22) = P(z >-2094) = P(z < 2.094) = 0.9817 (using z distribution table)

rounding the probability to 3 decimal places, we get 0.982

So, correct answer is option D

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