According to a survey in a country, 25% of adults do not have any credit cards.

Question

According to a survey in a country, 25% of adults do not have any credit cards. Suppose a simple random sample of 700 adults is obtained. Determine the standard deviation of the sampling distribution of p In a random sample of 700 adults, what is the probability that less than 24% have no credit cards? Would it be unusual if a random sample of 700 adults results in 196 or more having no credit cards? The result is unusual because the probability that p is greater than or equal to this sample proportion Is greater than 5% The result is unusual because the probability that p is greater than or equal to this sample proportion is less than 5% The result is not unusual because the probability that p is greater than w equal to this sample proportion is greater than 5%. The result is not unusual because the probability that p is greater than or equal to this sample proportion is less than 5%.

Explanation / Answer

Normal Approximation to Binomial Distribution
a.
Mean ( np ) =700 * 0.25 = 175
Standard Deviation ( npq )= 700*0.25*0.75 = 11.4564

b.
Proportion ( P ) =0.25
Standard Deviation ( sd )= Sqrt (P*Q /n) = Sqrt(0.25*0.75/700)
Normal Distribution = Z= X- u / sd ~ N(0,1)                  

P(X < 0.24) = (0.24-0.25)/0.0164
= -0.01/0.0164= -0.6098
= P ( Z <-0.6098) From Standard Normal Table
= 0.271                  

c.
P(X > 196) = (196-175)/11.4564
= 21/11.4564 = 1.833
= P ( Z >1.833) From Standard Normal Table
= 0.0334                  
d.
the result is unusual because the probability that p is greater than
equal to this sample proportion is less than 5%

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