Question

Tutorial Exercise In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal distribution to estimate the requested probabilities. Do you take the free samples offered in supermarkets? About 63% of all customers will take free samples. Furthermore, of those who take the free samples, about 32% will buy what they have sampled. Suppose you set up a counter in a supermarket offering free samples of a new product. The day you were offering free samples, 307 customers passed by your counter. (a) What is the probability that more than 180 will take your free sample? (b) What is the probability that fewer than 200 will take your free sample? (c) What is the probability that a customer will take a free sample and buy the product? Hint: Use the multiplication rule for dependent events. Notice that we are given the conditional probability P(buy sample) -0.32, while P(sample) = 0.63. (d) What is the probability that between 60 and 80 customers will take the free sample and buy the product? Hint: Use the probability of success calculated in part (c). Step 1 (a) What is the probability that more than 180 will take your free sample? We are asked to find the probability that more than 180 people of the 307 that walk by will take a free sample in a supermarket. We are told that about 63% of all customers take free samples. First we must test to see if we can use the normal approximation to the binomial distribution. Here we will define success as "someone takes a free sample." In this scenario, the number of trials is n . The probability of success is p , which means that the probability of a failure, 9, is 9-1-p- Since np and nq = 113.59, we ---Select- distribution because these values are both greater than 5. use the normal approximation to the binomial Submit Skip (you cannot come back)

Answer 1

here, n = number of customers = 307

p = the probability of success = 0.63

q= (1-p) = (1- 0.63)= 0.37

checking for normal approximation:-

np = (307*0.63) = 193.41

nq = (307*0.37) = 113.59

both np and nq are > 5, so we can use the normal approximation to the binomial distribution.

so, we can say that,

$X\sim N(\mu,\sigma)$

where, $\mu = np = 193.41$

$\sigma=\sqrt{npq}=\sqrt{325*0.63*0.37}=8.4594$

a).the probability that more than 180 people will take my free sample be:-

$P(X>180)$

$=P(X\geq180.5)$ [ continuity correction ]

$=P(\frac{X-193.41}{8.4594}\geq\frac{180.5-193.41}{8.4594})$

$=P(z\geq-1.53)$

$=1-\Phi(-1.53)$

$=\Phi(1.53)$

= 0.937 [ from standard normal table ]

b). the probability that fewer than 200 will take your sample be:-

$P(X<200)$

$=P(X\leq199.5)$ [ continuity correction ]

$=P(\frac{X-193.41}{8.4594}\leq\frac{199.5-193.41}{8.4594})$

$=P(z\leq0.72)$

$=\Phi(0.72)$

= 0.7642 [ from standard normal table ]

c). P(sample) = 0.63

$P(buy|sample)=0.32$

$\Rightarrow \frac{P(buy\cap sample)}{P(sample)}=0.32$

$\Rightarrow P(buy\cap sample)=0.32*0.63=0.2016$

the probability that a customer will take a free sample and buy the product = 0.2016

d). n = number of customers = 307

p = probability that a customer will take a free sample and buy the product = 0.2016

q= (1-p) = (1- 0.2016)= 0.7984

checking for normal approximation:-

np = (307*0.2016) = 61.8912

nq = (325*0.7984) = 245.1080

both np and nq are > 5, so we can do the normal approximation.

so, we can say that,

$X\sim N(\mu,\sigma)$

where, $\mu = np = 61.8912$

$\sigma=\sqrt{npq}=\sqrt{325*0.2016*0.7984}=7.0295$

the probability that between 60 and 80 customers will take free sample and buy the product be:-

$P(60<X<80)$

$=P(59.5\leq X\leq80.5)$[ continuity correction ]

$=P(\frac{59.5-61.8912}{7.0295}\leq \frac{X-61.8912}{7.0295} \leq\frac{80.5-61.8912}{7.0295})$

$=P(-0.34\leq z\leq 2.65)$

$=\Phi(2.65)-\Phi(-0.34)$

$=\Phi(2.65)&plus;\Phi(0.34)-1$

$=0.9960&plus;0.6331-1$ [ from standard normal table ]

= 0.6291

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