In order to avoid power failures due to falling limbs and down trees, power companies will intentionally kill trees growing near their lines. Applying a chemical to slow the growth of the trees is cheaper than trimming, but the chemical kills some of the trees. Suppose that one such chemical would kill 25% of elm trees. The power company tests the chemical on 300 elms. Consider these an SRS from the population of all elm trees. A) What are the mean and standard deviation of the proportion of trees that are killed?
B) Is it okay to use normal calculations for this problem? Explain.
C) What is the probability that at least 26% of the sample is killed?

a)

Sample mean p = 25% = 0.25

standard deviation σ = 0.025

b)

YES, it okay to use normal calculations for this problem since the sample size is 300, hence np>10 and nq>10

so the conditions satisfied for normal approximation/calculations

c) the probability that at least 26% of the sample is killed is 0.6554

Step-by-step explanation:

Given the data in the question;

Suppose that one such chemical would kill 25% of elm trees. The power company tests the chemical on 300 elms

sample size n = 300

a) What are the mean and standard deviation of the proportion of trees that are killed?

Sample mean p = 25% = 0.25

standard deviation σ = √( pq / n ) = √( (0.25 × (1 - 0.25)) / 300 )

σ = √( (0.25 × 0.75) / 300 )

σ = √( 0.1875 / 300 )

σ = √0.000625

σ = 0.025

standard deviation σ = 0.025

b) Is it okay to use normal calculations for this problem?

YES, it okay to use normal calculations for this problem since the sample size is 300, hence np>10 and nq>10

so the conditions satisfied for normal approximation/calculations

c) the probability that at least 26% of the sample is killed

Now we are to find

P( P < 0.26 )

since its a normal approximation, we can convert p to z

so

P( P < 0.26 ) =  P( Z < ( x-p / σ   )

P( P < 0.26 ) = P( Z < ( 0.26-0.25 / 0.025 )

= P( Z < (0.01 / 0.025) )

= P( Z < 0.4 )

form z score table; P( Z < 0.4 ) = 0.6554

so

P( P < 0.26 ) = 0.6554

the probability that at least 26% of the sample is killed is 0.6554

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step-by-step explanation: