Question 1022199: About 26% of ice-cream sales are vanilla. 14% account for chocolate sales. Suppose 184 customers go to a grocery store.
a)what is the probably that 50 or more will buy vanilla?
b) what is the probability 12 or more will buy chocolate
c) What is the probability that someone who is buying ice cream will buy chocolate or vanilla?
d) What is the probability that between 50 and 60 customers will buy chocolate or vanilla ice cream?
I'm so stuck on how to set up the problem.
Elin
Answer by mathmate(429)   (Show Source):
You can put this solution on YOUR website!
Question:
About 26% of ice-cream sales are vanilla. 14% account for chocolate sales. Suppose 184 customers go to a grocery store.
a)what is the probably that 50 or more will buy vanilla?
b) what is the probability 12 or more will buy chocolate
c) What is the probability that someone who is buying ice cream will buy chocolate or vanilla?
d) What is the probability that between 50 and 60 customers will buy chocolate or vanilla ice cream?
Setting up and example of solution:
Here we have a case where:
1. The number of trials (n=184) is known.
2. Each trial is a Bernoulli experiment (true or false, e.g. vanilla or not vanilla, chocolate, or not cholate).
3. All trials are (assumed) random and independent of each other.
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).
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:
P(k;n;p)= .............(1)
where
C(n,k)=n!/(k!(n-k)!) is coefficient of combination of k objects out of n.
(a)
So we need the probability of k>=50 customers would buy vanilla.
We need
P(k>=50;n;p)
=sum of all cases of P(k;n;p) for k=50,51,....184.
OR
= 1-sum of all cases of P(k;n;p) for k=0,1,2,3....49.
This means a lot of calculations using equation (1).
You would probably need a scientific calculator with the binomial distribution built-in, or an application such as R.
The latter gives
P(k>=50;184;0.26)=0.3854 [pbinom(49,184,0.26]
If you don't have an advanced scientific calculator, you could use, in this particular case the normal approximation to the binomial, which uses
mean=μ=np=184*0.26=47.84,
standard deviation=σ=sqrt(variance)=sqrt(np(1-p))=sqrt(184*.26*.74)=5.95
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).
Now we need to convert k to Z (for normal distribution),
Zmin=(Xmin-μ)/σ=(49.5-47.84)/5.95=0.27899
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).
You can proceed with the other parts similarly.
Feel free to post your answer for checking.
|
|
|
