Question

(1) American ladybugs have an average adult length of 1 cm with a known standard deviation of 0.2 cm. The population of American ladybugs in Raleigh was around 1914000 last spring. Assume a normal distribution for the lengths of adult American ladybugs.

Your niece asks you what's the probability of a random ladybug in Raleigh being bigger than 1.5 cm. Is it appropriate to calculate this probability?

Select one: a. Yes.

b. No, because the population distribution is skewed.

c. No, because the sample size is less than 30.

d. No, because the empirical rule is violated.

(2)Regardless of your answer to the previous question, calculate this probability using a normal distribution. Report your answer to four decimal places.

(3)

Although it would be difficult in practice, assume we are able to randomly sample 20 American ladybugs. What's the sampling distribution of the sample mean of these ladybugs?

Select one:

a. A t-distribution with 19 degrees of freedom

b. The sampling distribution is unknown because relevant assumptions are violated.

c. A normal distribution with mean 1 and standard deviation 0.2.

d. A normal distribution with mean 1 and standard deviation 0.045.

(4) Calculate the probability of observing an average American ladybug length between 0.95 cm and 1.05 cm for a random sample of 20 ladybugs. Give your answer accurate to four decimal places. If you found assumptions to be violated in the previous question, answer this question as if the assumptions had not been violated.

Answer 1

Answer:

1)

a. Yes.

The sampling distribution of any sample size will be approximately normally distributed if the population distribution is normal. We can use normal distribution to calculate the probability.

2)

P(X > 1.5) = P[Z > (1.5 - 1)/0.2]  

= P[Z > 2.5]

= 0.0062

3)

By Central limit theorem, the sampling distribution of the sample mean will be normal distribution with mean = 1 cm and standard deviation = 0.2/ = 0.045

Option D is Correct

d. A normal distribution with mean 1 and standard deviation 0.045.

4)

P(0.95 < < 1.05) = P( < 1.05) - P(< 0.95)

= P[Z < (1.05 - 1)/0.045] - P[Z < (0.95 - 1)/0.045]

= P[Z < 1.11] - P[Z < -1.11]

= 0.8665 -  0.1335

= 0.7330

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