For each, find the missing value using an appropriate continuity correction.

$P_{\textrm{binomial}}(x \ge 3) = P_{\textrm{normal}}(x \gt ?)$

$P_{\textrm{binomial}}(x \le 3) = P_{\textrm{normal}}(x \lt ?)$

$P_{\textrm{binomial}}(x \gt 3) = P_{\textrm{normal}}(x \gt ?)$

$P_{\textrm{binomial}}(x \lt 3) = P_{\textrm{normal}}(x \lt ?)$

- $P_b(3) + P_b(4) + P_b(5) + \cdots + P_b(N) \approx P_n(x \gt 2.5) $
- $P_b(0) + P_b(1) + P_b(2) + P_b(3) \approx P_n(-0.5 \lt x \lt 3.5) \approx P_n(x \lt 3.5)$
- $P_b(4) + P_b(5) + P_b(6) + \cdots + P_b(N) \approx P_n(x \gt 3.5)$
- $P_b(0) + P_b(1) + P_b(2) \approx P_n(-0.5 \lt x \lt 2.5) \approx P_n(x \lt 2.5)$

Use the normal approximation to the binomial with $n = 30$ and $p = 0.5$ to find the probability $P(X = 18)$.

$n=30$ and $p=0.5$, so we first check to see if the normal approximation is appropriate. $np = 15 \ge 5$ and $nq = 15 \ge 5$, so it is. Then for the approximating normal distribution, $\mu = np = 15$, $\sigma = \sqrt{npq} = 2.7386$. Now note that the binomial probability we seek is $P(18)$ which is approximated by the normal probability $P(17.5 \lt x \lt 18.5)$. Thus, we find $z_{17.5} = 0.9129$ and $z_{18.5} = 1.2780$ and the corresponding probability $P(0.9129 \lt z \lt 1.2780) = 0.0800$ is our answer.Use the normal approximation to the binomial with $n = 10$ and $p = 0.5$ to find the probability $P(X \ge 7)$.

$n=10$ and $p=0.5$, so we first check to see if the normal approximation is appropriate. $np = 5 \ge 5$, and $nq = 5 \ge 5$, so it is. Then for the approximating normal distribution, $\mu = np = 5$ and $\sigma = \sqrt{npq} = 1.5811$. The binomial probability sought, $P(x \ge 7)$ is approximated by the normal probability $P(6.5 \lt x)$, so we find $z_{6.5} = 0.9487$. The related probability $P(z \gt 0.9487) = 1 - P(z \lt 0.9487) = 0.8286$ is our answer.Use the normal approximation to the binomial with $n = 50$ and $p = 0.6$ to find the probability $P(X \le 40)$.

$n=50$ and $p=0.6$, so we first check to see if the normal approximation is appropriate. $np = 30 \ge 5$ and $nq = 20 \ge 5$, so it is. Then for the approximating normal distribution, $\mu = np = 30$ and $\sigma = \sqrt{npq} = 3.464$. The binomial probability sought, $P(x \le 40)$ is approximated by the normal probability $P(x \lt 40.5)$, so we find $z_{40.5} = 3.0311$. The related probability $P(z \lt 3.0311) = 0.9988$ is our answer.According to recent surveys, 53% of households have personal computers. If a random sample of 175 households is selected, what is the probability that more than 75 but fewer than 110 have a personal computer?

$n=175$ and $p=0.53$, so we first check to see if the normal approximation is appropriate. $np = 92.75 \ge 5$ and $nq = 82.25 \ge 5$, so it is. Then for the approximating normal distribution, $\mu = np = 92.75$ and $\sigma = \sqrt{npq} = 6.6025$. The binomial probability sought, $P(75 \lt x \lt 110)$ is approximated by the normal probability $P(75.5 \lt x \lt 109.5)$, so we find $z_{75.5} = -2.6127$ and $z_{109.5} = 2.5369$. The related probability $P(-2.6127 \lt z \lt 2.5369) = 0.9899$ is our answer.According to Mars, 24% of M&M plain candies are blue. In a given sample of 100 M&Ms, 27 are found to be blue. Assuming that the claimed rate of 24% is correct, find the probability of randomly selecting 100 M&Ms and getting 27 or more that are blue. Based on the result, is 27 (out of 100) an unusually high number of blue M&Ms?

$n=100$ and $p=0.24$, so we first check to see if the normal approximation is appropriate. $np = 24 \ge 5$ and $nq = 76 \ge 5$, so it is. Then for the approximating normal distribution, $\mu = np = 24$ and $\sigma = \sqrt{npq} = 4.2708$. The binomial probability sought, $P(27 \le x)$ is approximated by the normal probability $P(26.5 \lt x)$, so we find $z_{26.5} = 0.5854$. The related probability $P(0.5854 \lt z) = 0.2791$ is our answer.A Boeing 767-300 aircraft has 213 seats. When someone buys a ticket for a flight there is a 0.0995 probability that the person will not show up for the flight. A ticket agent accepts 236 reservations for a flight that uses a Boeing 767-300. Find the probability that not enough seats will be available. Is this probability low enough so that overbooking is not a real concern? If not, how many tickets should be sold so that the probability is less than 10% that at least one person will not have a seat?

$n=236$ and $p=1-0.0995=0.9005$, where "success" is someone shows up for their flight. We first check to see if the normal approximation is appropriate. $np = 212.518 \ge 5$ and $nq = 23.482 \ge 5$, so it is. Then for the approximating normal distribution, $\mu = np = 212.518$ and $\sigma = \sqrt{npq} = 4.5984$. Not enough seats are available when more than 213 people show up, so we seek the binomial probability $P(x > 213)$ which is approximated by the normal probability $P(x \gt 213.5)$, so we find $z_{213.5} = 0.2136$. The related probability $P(z \gt 0.2136) = 0.4154$ gives the probability that not enough seats will be available. This does not seem to be an insignificant probability. If we want the probability that at least one person will not have a seat to be less than $0.10$, then we need to limit the number of accepted reservations to $230$.Let $X$ be the random variable that represents a count of the number of heads showing when a coin is tossed 12 times. Find the following binomial probabilities exactly, and then compute their corresponding normal approximations: (a) P(at least 8 heads), (b) P(exactly 5 heads), (c) P(at most 5 heads)

- binomial 0.1938; normal 0.1932
- binomial 0.1934; normal 0.1932
- binomial 0.3872; normal 0.3863

Toss a coin 16 times. Let X be the number of heads that appear. Find the probability that there will be more than 13 heads. (a) Work as a binomial. (b) Use the standard normal to approximate the probability. (c) Was the approximation reasonable? Explain.

(a) 0.0021, (b) 0.0030, (c) Yes, the approximation is reasonable (difference was less than $0.001$). We could have predicted this as $np \ge 5$ and $nq \ge 5$.Assume that the probability of a college student having a car on campus is .30. A random sample of 12 students is taken. What is the probability that at least 4 will have a car on campus? (a) Work the problem as a binomial. (b) Approximate the probability using the standard normal. (c) Is the approximation reasonable? Explain clearly.

(a) 0.5075, (b) 0.5251, (c) No, the approximation is not reasonable - it is off by almost 2 percent. We could have predicted this as as $np = 3.6 \lt 5$.Previous studies have found that $75\%$ of adults use the internet on a regular basis. A researcher believes this percentage has recently increased. He conducts a survey and discovers that 2144 out of 2824 adults surveyed use the internet on a regular basis. Assuming $75%$ actually is the correct percentage, what's the probability of seeing at least this many users out of 2824 users? Does this happening seem to provide evidence that the percentage of adult internet users has increased?

This is binomial with $n = 2824$, $p = 0.75$, $q = 0.25$. Noting that $np,nq \ge 5$, approximating with a normal distribution is appropriate. We seek $P_{binomial}(x \ge 2144) \approx P_{normal}(x \ge 2143.5)$. $\mu = 2118$ and $\sigma \doteq 23.01$, so $z = 1.1082$ for $x=2143.5$. Thus, we find $P_{std norm}(z \ge 1.1082) \doteq 0.1339$. This is not that unlikely. By itself, it doesn't seem that we have compelling evidence that the percentaage of adult internet users has increased.Benford's law observes that in many real-life sets of numerical data, the leading significant digit is not uniformly distributed between $1$ through $9$. Instead, the probability that the leading digit is $d$ is given by $$P(d) = \log_{10} \left(1 + \frac{1}{d}\right)$$ It is known that check amounts issued by companies generally follow Benford's law. An investigator analyzes 784 randomly selected checks issued by a company suspected of fraud. Of these, 479 of them were issued for amounts beginning with the digit 5. According to the formula above, our expectation is that this percentage should be 7.9%. Assume that Benford's law applies in this situation. What is the probability of seeing 479 or more amounts starting with a 5? Does this probability suggest this is a rare or common event? What should one conclude as a result?

This is binomial with $n = 784$, $p = 0.079$, $q = 0.921$. Noting that $np,nq \ge 5$, approximating with a normal distribution is appropriate. We seek $P_{binomial}(x \ge 479) \approx P_{normal}(x \ge 478.5)$. $\mu = 61.936$ and $\sigma \doteq 7.5527$, so $z \doteq 55.15$ for $x = 478.5$. Thus we find $P_{std norm}(z \ge 55.15)$, which is so small it registers as $0$ on most calculators. This is a ridiculously rare event and suggests in the strongest way possible that something (possibly fraud) is keeping the distribution in question from following Benford's law.Six percent of people are universal blood donors (i.e., they can give blood to anyone without it being rejected). A hosptial needs 10 universal donors to donote blood, so they conduct a blood drive. If 200 volunteers donate blood, what is the probability tht the number of universal donors is at least 10? Is the pool of 200 volunteers likely to be sufficient?

This is binomial with $n = 200$, $p = 0.06$, $q = 0.94$. Noting that $np,nq \ge 5$, approximating with a normal distribution is appropriate. We seek $P_{binomial}(x \ge 10) \approx P_{normal}(x \ge 9.5)$. $\mu = 12$ and $\sigma \doteq 3.3586$, so $z \doteq -0.7444$ for $x = 9.5$. Thus we find $P_{std norm}(z \ge -.7444) \doteq 0.7717$. So there is slightly more than a 22% chance that the blood drive does not produce enough universal donors.