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ELEN [110]
2 years ago
6

Only 35% of the drivers in a particular city wear seat belts. Suppose that 20 drivers are stopped at random what is the probabil

ity that exactly four are wearing a seatbelt? (Round your answer to 4 decimal places)
Mathematics
1 answer:
lesya692 [45]2 years ago
6 0

Here we have a situation where the probability of a driver wearing seat belts is known and remains constant throughout the experiment of stopping 20 drivers.

The drivers stopped are assumed to be random and independent.

These conditions are suitable for modelling using he binomial distribution, where

where n=number of drivers stopped (sample size = 20)

x=number of drivers wearing seatbelts (4)

p=probability that a driver wears seatbelts (0.35), and

C(n,x)=binomial coefficient of x objects chosen from n = n!/(x!(n-x)!)

So the probability of finding 4 drivers wearing seatbelts out of a sample of 20

P(4;20;0.35)

=C(20,4)*(0.35)^4*(0.65)^16

= 4845*0.0150061*0.0010153

= 0.07382

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Alternate Hypothesis, H_A : \mu > 210 seconds      {means that the sample is from a population of songs with a mean greater than 210 seconds}

The test statistics that will be used here is <u>One-sample z-test statistics </u>because we know about population standard deviation;

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The value of z-test statistics is 4.932.

Now, at 0.05 level of significance, the z table gives a critical value of 1.645 for the right-tailed test.

Since the value of our test statistics is more than the critical value of z as 4.932 > 1.645, so <u><em>we have sufficient evidence to reject our null hypothesis</em></u> as it will fall in the rejection region.

Therefore, we conclude that the sample is from a population of songs with a mean greater than 210 seconds.

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