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kherson [118]
3 years ago
13

A market research firm knows from historical data that telephone surveys have a 36% response rate. In a random sample of 280 tel

ephone numbers, what is the probability that the response rate will be between 33.5% and 39%?
Mathematics
2 answers:
AVprozaik [17]3 years ago
5 0

Answer:

Probability that the response rate will be between 33.5% and 39% = 0.66176 .

Step-by-step explanation:

We are given that a market research firm knows from historical data that telephone surveys have a 36% response rate.

The probability criterion we will use here is;

           \frac{\hat p- p}{\sqrt{\frac{\hat p (1-\hat p)}{n} } } ~ N(0,1)

Here, p = 0.36 and n = sample size = 280

Let \hat p = response rate

So, P(0.335 <= \hat p <= 0.39) = P(\hat p <= 0.39) - P(\hat p < 0.335)

P(\hat p <= 0.39) = P( \frac{\hat p- p}{\sqrt{\frac{\hat p (1-\hat p)}{n} } } <= \frac{0.39- 0.36}{\sqrt{\frac{0.39 (1-0.39)}{280} } } ) = P(Z <= 1.03) = 0.84849

P(\hat p < 0.335) = P( \frac{\hat p- p}{\sqrt{\frac{\hat p (1-\hat p)}{n} } } < \frac{0.335- 0.36}{\sqrt{\frac{0.335 (1-0.335)}{280} } } ) = P(Z < -0.89) = 1 - P(Z <= 0.89)

                                                                  = 1 - 0.81327 = 0.18673

Therefore, P(0.335 <= \hat p <= 0.39) = 0.84849 - 0.18673 = 0.66176

Hence, probability that the response rate will be between 33.5% and 39% is 0.66176 or 66.18 % .

Rasek [7]3 years ago
3 0

Answer:

0.6604

Step-by-step explanation:

Given that a market  research firm knows from historical data that telephone surveys have a 36% response rate.

Sample size of random sample = 280

We know for samples randomly drawn of large size sample proportion follows a normal distribution with mean= sample proportin and std error

= \sqrt{\frac{pq}{n} }

Substitute p = 0.36 and q = 1-0.36= 0.64

p follows N with mean = 0.36 and std dev = \sqrt{\frac{0.36*0.64}{\sqrt{280} } } \\=0.0287

Using normal distribution values we can find\

P(33.5p.c. < p < 39pc)\\= P(0.335

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