Question 1)A binomial experiment is given. Decide whether you can use the normal distribution to approximate the binomial distribution. If you? can, find the mean and standard deviation. If you? cannot, explain why. A survey of adults found that 47?% have used a multivitamin in the past 12 months. You randomly select 50 adults and ask them if they have used a multivitamin in the past 12 months. What is the mean and standard deviation? (Round to two decimal places)

 

Question 2) Five percent of workers in a city use public transportation to get to work. A transit authority offers discount rates to companies that have at least 30 employees who use public transportation to get to work. You randomly select 3 companies and ask their employees if they use public transportation below. Complete parts? (a) through? (c) below. ?(a) Company A has 280 employees. What is the probability that Company A will get the? discount? (Round to 4 decimal places as needed)

 

Question 3) A drug tester claims that a drug cures a rare skin disease 71?% of the time. The claim is checked by testing the drug on 100 patients. If at least 69 patients are? cured, the claim will be accepted. Find the probability that the claim will be rejected assuming that the? manufacturer's claim is true. Use the normal distribution to approximate the binomial distribution if possible. The probability is? (Round to four decimal places)

Please read the explanation for complete answer.

Step-by-step explanation

1.

The general rule of thumb is that the sample size n is "sufficiently large" for normal approximation of binomial if:

np≥5  and  n(1−p)≥5.

If these conditions are satisfied, then we say that the given binomial random variable follows approximately a normal distribution with

mean, μ=np  and  variance, σ2=np(1−p).

We have, n = 50, p = 0.47. So,

np=50∗0.47=23.5≥5n(1−p)=50∗0.53=26.5≥5.

Since, the required conditions are met, we can use the normal approximation.

Mean, μ=np=50∗0.47=23.5Standard deviation=variance?=np(1−p)?=50∗0.47∗0.53?=3.53

2.

Company A will get the? discount if it has at least 30 employees who use public transportation to get to work.

Given that, number of employees at Company A, n = 280.

5% of workers in the city use public transportation to get to work. Hence, p = 0.05.

Let X denotes the number of employees at Company A those who use public transportation to get to work. Then X follows a binomial distribution with n = 280 and p = 0.05.

We have to find the probability, P(X is at least 30).

Let us check if we can use normal approximation.

np=280∗0.05=14≥5n(1−p)=280∗0.95=266≥5

Since, the required conditions are met, we can use the normal approximation.

Mean, μ=np=280∗0.05=14Variance, σ2=np(1−p)=280∗0.05∗0.95=13.3

P(X is at least 30)=P(X≥30)=P(σX−μ?≥σ30−μ?)=P(13.3?X−14?≥13.3?30−14?)=P(Z≥4.387)=5.74×10−6≈0

{Excel-2019 formula to calculate this probability is =1-NORM.S.DIST(4.387,TRUE) }

Hence, probability that Company A will get the? discount is almost zero.

{NOTE: If we don't use the normal approximation, then

P(X≥30)=0.000084≈0.0001

You can get this in Microsoft Excel-2019 by running the formula =1-BINOM.DIST(29,280,0.05,TRUE) }

3.

Number of patients tested, n = 100.

Assuming that the? manufacturer's claim is true, the probability that the drug cures a patient is, p = 0.71.

Let X denotes the number of patients cured. Then X follows a binomial distribution with n = 100 and p = 0.71.

np=100∗0.71=71≥5n(1−p)=100∗0.29=29≥5

Since, the required conditions are met, we can use the normal approximation.

Mean, μ=np=100∗0.71=71Variance, σ2=np(1−p)=100∗0.71∗0.29=20.59

If at least 69 patients are? cured, the claim will be accepted. We have to find the probability that the claim will be rejected, i.e; the probability that less than 69 patients are? cured.

P(X<69)=P(σX−μ?<σ69−μ?)=P(20.59?X−71?<20.59?69−71?)=P(Z<−0.4408)=0.3297

Hence, probability that the claim will be rejected is 0.3297.

{Excel-2019 formula to calculate this probability is =NORM.S.DIST(-0.4408,TRUE) }

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