Question

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)

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<X - U/S)

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.

Answer 2

The true statements are:

• The mean is 32, and the standard error is 0.068
• The expected shape will be approximately normal
• The probability of scoring less than 60% is 0.278

The given parameters are:

n = 50 --- number of questions

p = 0.64 -- the probability of getting the correct answer

The mean is calculated as:

$E(x) = np$

So, we have:

$E(x) = 50 * 0.64$

$E(x) = 32$

The standard error is calculated as:

$SE= \sqrt{\frac{p * (1 - p)}{n}}$

So, we have:

$SE= \sqrt{\frac{0.64 * (1 - 0.64)}{50}}$

$SE= \sqrt{0.004608}$

$SE= 0.068$

Hence, the mean is 32, and the standard error is 0.068

The sample size is large, so the expected shape will be approximately normal

To calculate the probability of scoring less than 60%, we make start by calculating the z-score using:

$z = \frac{x - \mu}{\sigma}$

So, we have:

$z = \frac{0.60 - 0.64}{0.068}$

$z = -0.588$

So, we have:

$P(x < 60\%) = P(z < -0.588)$

$P(x < 60\%) = 0.278$

Hence, the probability of scoring less than 60% is 0.278

Read more about normal distributions at:

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