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3 years ago

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Sixty-eight students, out of 100 surveyed, chose pizza as their favorite lunch item. If the school has 1250 students, how many s

tudents would likely say that pizza is their favorite if the survey is a fair representation of the student body?

1 answer:

7nadin3 [17]3 years ago

3 0

Answer:

850 students would say that pizza is their favorite

Step-by-step explanation:

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90 is 120% of what number?

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The probability for event A is 0.3, the probability for event B is 0.6, and the probability of events A or B is 0.8. Why are the

Sedaia [141]

Answer:

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Step-by-step explanation:

I think it makes the most sense sorry if i got it wrong

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3 years ago

Manufacturers are testing a die to make sure that it is fair (has a uniform distribution). They roll the die 66 times and record

steposvetlana [31]

Answer:

We should reject H0

At the 5% significance level, there is sufficient evidence to conclude that the die is not fair.

Step-by-step explanation:

Null hypothesis is a statement that is to be tested against the alternative hypothesis and then decision is taken whether to accept or reject the null hypothesis. The critical value is 15.091 and test statistic is 11.070. The null hypothesis is rejected or accepted on the basis of level of significance. When the p-value Test statistics is greater than level of significance we fail to reject the null hypothesis and null hypothesis is then accepted. In the given case p-value is less than critical value then we should reject the null hypothesis.

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3 years ago

A new medical test provides a false positive result for hepatitis 2% of the time that is a perfectly healthy subject being teste

Oxana [17]

Given that X <span>be the number of subjects who test positive for the disease out of the 30 healthy subjects used for the test.

The probability of success, i.e. the probability that a healthy subject tests positive is given as 2% = 0.02

Part A:

</span><span>The probability that all 30 subjects will appropriately test as not being infected, that is the probability that none of the healthy subjects will test positive is given by:

</span> $P(X)={ ^nC_xp^x(1-p)^{n-x}} \\ \\ P(0)={ ^{30}C_0(0.02)^0(1-0.02)^{30-0}} \\ \\ =1(1)(0.98)^{30}=0.5455$
<span>

Part B:

The mean of a binomial distribution is given by

</span> $\mu=np \\ \\ =30(0.02) \\ \\ =0.6$
<span>
The standard deviation is given by:

</span> $\sigma=\sqrt{np(1-p)} \\ \\ =\sqrt{30(0.02)(1-0.02)} \\ \\ =\sqrt{30(0.02)(0.98)}=\sqrt{0.588} \\ \\ =0.7668$
<span>

Part C:

This test will not be a trusted test in the field of medicine as it has a standard deviation higher than the mean. The testing method will not be consistent in determining the infection of hepatitis.</span>

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3 years ago