An exam consists of 50 multiple choice questions. Based on how much you studied, for any given question, you think you have a probability of 0.64 of getting the correct answer. Consider the sampling distribution of the sample proportion of correct questions out of 50. (a) Find the mean and standard error of the sampling distribution of this proportion. Mean = (2 decimal places) Standard error = (3 decimal places) (b) What do you expect for the shape of this sampling distribution? approximately normal because n is large cannot be determined because p is greater than 0.5 cannot be determined because n is large approximately normal because the population is normal (c) If truly p = 0.64, calculate the probability of getting a sample proportion less than 0.60? (correct answers on less than 60% of the questions) Probability = (3 decimal places)

Question

Rosemary

Mathematics

19 September, 23:21

An exam consists of 50 multiple choice questions. Based on how much you studied, for any given question, you think you have a probability of 0.64 of getting the correct answer. Consider the sampling distribution of the sample proportion of correct questions out of 50. (a) Find the mean and standard error of the sampling distribution of this proportion. Mean = (2 decimal places) Standard error = (3 decimal places) (b) What do you expect for the shape of this sampling distribution? approximately normal because n is large cannot be determined because p is greater than 0.5 cannot be determined because n is large approximately normal because the population is normal (c) If truly p = 0.64, calculate the probability of getting a sample proportion less than 0.60? (correct answers on less than 60% of the questions) Probability = (3 decimal places)

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Answers (1)

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Jay Dickson · Answer 1

Answer: (a) mean = 32.00

(b) Standard error = 0.068

It is expected that the shape will approximately large because n is large.

(c) P (z< 0.60) = 0

Step-by-step explanation:

a. Mean = nP = 0.64 * 50 = 32.00 (2dp)

P = 0.64

b. Standard error = √ p (1-p) / n

= √0.64 (0.36) / 50

= 0.068 (3dp)

c. Using P (Z
Where S = standard error

U = mean = 0.64

(c) P (z< 0.60) =

= P (z< 0.60 - 0.64/0.068)

= P (z < - 5.88)

= 0.