SOLUTION: In a survey of adults throughout America 75% strongly support integration.You randomly select 24 adults ,throughout America and ask each if they strongly support integration.
a/
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Question 363377: In a survey of adults throughout America 75% strongly support integration.You randomly select 24 adults ,throughout America and ask each if they strongly support integration.
a/ Show the the conditions for a binomial distribution are met
b/ Decide whether you can use the normal distribution to approximate the binomial distribution.
c/Find the Probability that 15 adults say they strongly support integration
Answer by edjones(8007) (Show Source): You can put this solution on YOUR website!
a. People either do or don't strongly support integration.
.
b. A normal distribution will not work here. this is a 75-25 proposition.
.
c. Let x=.75 and y=.25
24C15=1307504*x^15*y^9
=.0667 Probability that EXACTLY 15 adults say they strongly support integration.
.
Ed
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