document.write( "Question 1022199: About 26% of ice-cream sales are vanilla. 14% account for chocolate sales. Suppose 184 customers go to a grocery store.
\n" ); document.write( "a)what is the probably that 50 or more will buy vanilla?
\n" ); document.write( "b) what is the probability 12 or more will buy chocolate
\n" ); document.write( "c) What is the probability that someone who is buying ice cream will buy chocolate or vanilla?
\n" ); document.write( "d) What is the probability that between 50 and 60 customers will buy chocolate or vanilla ice cream?\r
\n" ); document.write( "\n" ); document.write( "I'm so stuck on how to set up the problem.
\n" ); document.write( "Elin
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Algebra.Com's Answer #637988 by mathmate(429)\"\" \"About 
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\n" ); document.write( "Question:
\n" ); document.write( "About 26% of ice-cream sales are vanilla. 14% account for chocolate sales. Suppose 184 customers go to a grocery store.
\n" ); document.write( "a)what is the probably that 50 or more will buy vanilla?
\n" ); document.write( "b) what is the probability 12 or more will buy chocolate
\n" ); document.write( "c) What is the probability that someone who is buying ice cream will buy chocolate or vanilla?
\n" ); document.write( "d) What is the probability that between 50 and 60 customers will buy chocolate or vanilla ice cream?
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\n" ); document.write( "Setting up and example of solution:
\n" ); document.write( "Here we have a case where:
\n" ); document.write( "1. The number of trials (n=184) is known.
\n" ); document.write( "2. Each trial is a Bernoulli experiment (true or false, e.g. vanilla or not vanilla, chocolate, or not cholate).
\n" ); document.write( "3. All trials are (assumed) random and independent of each other.
\n" ); document.write( "4. The probability of each outcome is constant throughout the experiment (0.26 for vanilla, 0.14 for chocolate, meaning 0.40 for either one of the two).
\n" ); document.write( "Under these circumstances, we can model the situation with a binomial distribution, which says that the probability of k successes out of n trials each with a probability of success p is given by:
\n" ); document.write( "P(k;n;p)=\"C%28n%2Ck%29%2Ap%5Ek%2A%281-p%29%5E%28n-k%29\".............(1)
\n" ); document.write( "where
\n" ); document.write( "C(n,k)=n!/(k!(n-k)!) is coefficient of combination of k objects out of n.
\n" ); document.write( "
\n" ); document.write( "(a)
\n" ); document.write( "So we need the probability of k>=50 customers would buy vanilla.
\n" ); document.write( "We need
\n" ); document.write( "P(k>=50;n;p)
\n" ); document.write( "=sum of all cases of P(k;n;p) for k=50,51,....184.
\n" ); document.write( "OR
\n" ); document.write( "= 1-sum of all cases of P(k;n;p) for k=0,1,2,3....49.
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\n" ); document.write( "This means a lot of calculations using equation (1).
\n" ); document.write( "You would probably need a scientific calculator with the binomial distribution built-in, or an application such as R.
\n" ); document.write( "The latter gives
\n" ); document.write( "P(k>=50;184;0.26)=0.3854 [pbinom(49,184,0.26]
\n" ); document.write( "
\n" ); document.write( "If you don't have an advanced scientific calculator, you could use, in this particular case the normal approximation to the binomial, which uses
\n" ); document.write( "mean=μ=np=184*0.26=47.84,
\n" ); document.write( "standard deviation=σ=sqrt(variance)=sqrt(np(1-p))=sqrt(184*.26*.74)=5.95
\n" ); document.write( "The required range is from 50 to 184, but since 49.5+ would round to 50, so we will calculate the range of 49.5<=k<=184 (called the continuity correction).
\n" ); document.write( "Now we need to convert k to Z (for normal distribution),
\n" ); document.write( "Zmin=(Xmin-μ)/σ=(49.5-47.84)/5.95=0.27899
\n" ); document.write( "We then look up the upper tail from a normal distribution table to get the probability of Z>=0.27900 (upper tail) to be 0.3901, not that far from the exact answer of 0.3854 (1.2% over).
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\n" ); document.write( "You can proceed with the other parts similarly.
\n" ); document.write( "Feel free to post your answer for checking.
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