In Exercises 5 and 6, determine whether you can use a normal distribution to app
ID: 3219536 • Letter: I
Question
In Exercises 5 and 6, determine whether you can use a normal distribution to approximate the binomial distribution. If you can, use the normal distribution to approximate the indicated probabilities and sketch their graphs. If you cannot, explain why and use a binomial distribution to find the indicated probabilities. 5. A survey of U.S. adults found that 64% watch NFL football. You randomly select 20 U.S. adults and ask them whether they watch NFL football. Find the probability that the number who watch NFL football is (a) exactly 10, (b) less than 7, and (c) at least 15, and (d) identify any unusual events. Explain. (Source: Harris Interactive) 6. A survey of U.S. adults ages 25 and older found that 86% have a high school diploma. You randomly select 30 U.S. adults ages 25 and older. Find the probability that the number who have a high school diploma is (a) exactly 25, (b) more than 25, and (c) less than 25, and (d) identify any unusual events. Explain. (Source: U.S. Census Bureau) In Exercises 7-12, use the following information. The amounts of time Facebook users spend on the website each month are normally distributed, with a mean of 6.7 hours and a standard deviation of 1.8 hours. (Adapted from Nielsen) 7. Find the probability that a Facebook user spends less than four hours on the website in a month. Is this an unusual event? Explain. 8. Find the probability that a Facebook user spends more than 10 hours on the website in a month. Is this an unusual event? Explain. 9. Out of 800 Facebook users, about how many would you expect to spend hours on the website in a month?Explanation / Answer
ANSWER :
5)
a)
Given n=20 p=0.64 q=0.36
P(10)=20c10 * 0.64^10 * 0.36^10 = 0.077
b)
Given n=20 p=0.64 q=0.36
P(0)=20c0 * 0.64^0 * 0.36^20 = 1.336e ^-9
P(1)=20c1 * 0.64^1 * 0.36^19 = -4.515e^-8
P(2)=20c2 * 0.64^2 * 0.36^18 = 4.224e^-8
P(3)=20c3 * 0.64^3 * 0.36^17 = 7.510e^-9
P(4) = 20c4 * 0.64^4 * 0.36^16 = -2.670e^-8
P(5)=20c5 * 0.64^5 * 0.36^15 = 5.160 e^-5
P(6)=20c6 * 0.64^6 * 0.36^14 = -4.220e^-8
P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(6) = 0.0001
C) atleast 15
Given n=20 p=0.64 q=0.36
P(15)=20c15 * 0.64^15 * 0.36^5 = 0.0162
P(16)=20c16 * 0.64^16 * 0.36^4 = -2.6614 e^-5
P(17)=20c17 * 0.64^17 * 0.36^3 = 2.3657 e^-5
P(18)=20c18 * 0.64^18 * 0.36^2 = 4.20 e^-5
p(19) = 20c19 * 0.64^19 * 0.36^1 = -0.0014
P(20)=20c20 * 0.64^20 * 0.36^0 = 1.3292
P(15)+P(16)+P(17)+P(18)+P(19)+P(20) = 0.015
7)
Given mean = 6.7 hours and standard deviation = 1.8 hours
Z = (Xbar - mu)/stdev
probability that a user spends less than 4 hours on the site a month = P[Z < (4 - 6.7)/1.8] = P[Z < -1.5] = 0.0668
8)
probability that a user spends more than ten hours on the site a month = 1 - P[Z < (10 - 6.7)/1. = 1 - P[Z < 1.83]
= 1 - 0.9666 = 0.0334