Question

Spring 2020 5. Approximately 14% of the world's population speaks Chinese as their native language. (a) If we randomly select 405 people from anywhere in the world, find the mean and the standard deviation for the count of Chinese speakers in the sample. (5 points) n = YUS P .147 (b) A sample of 405 people from all people is selected, determine whether finding 33 or more people who speak Chinese in this sample would be unusual. Use the z-score rule for unusual values to make your decision. (5 points) (c) A sample of 405 people is selected. Determine the probability of having 13 or fewer people who speak Chinese in this sample. Use the Z-score to get the probability Draw a normal curve and shade the area that represents the probability. (7 points)

Answer 1

Answer:

Given that:

Approximately 14% of the world,s population speaks Chinese as their native language

a) If we randomly select 405 people from any here in the world,find the mean and standard deviation for the count of Chinese speakers in the sample?

n = 405, p = 0.14

Mean = np

Mean = 405 * 0.14

Mean = 56.7

Standard deviation =

Vnp(1-p)

0= 405 * 0.14(1 – 0.14)

0= V 405 *0.14 * 0.86

0 = 48.762

6786'9 = 0

b) A sample of 405 people from all people is selected,determine whether finding 33 or more people who speak Chinese in this sample would be unusual

P(x > 33) = PX - 33 – 56.7 6.9829

P(X > 33) = P(Z > -23.7 6.9829'

P(X > 33) = P(Z > -2.44)

P(X > 33) = 1 - P(Z < -2.44)

P(X > 33) = 1-0.0073

P(X > 33) = 0.9927

Probability that 33 or more people who speak Chinese is 0.9927

0. 27 33.

c) A sample of 405 people is selected.Determine the probability of having 13 of fewer people who speak Chinese in this sample

P(X <33) = PX - 33 – 56.7, 6.9829

-23.7 P(X <33) = P(Z <a 6,9829

P(X <33) = P(Z < -2.44)

P(X <33) = 0.0073

Probability that 33 or fewer than speak Chinese is 0.0073

0.003 22