You may need to use the appropriate appendix table or technology to answer this question. Assume a binomial probability distribution has $$ p=0.60 $$ and $$ n=300 $$. (a) What are the mean and standard deviation? (Round your answers to two decimal places.) mean $$ \square $$ standard deviation $$ \square $$ (b) Is this situation one in which binomial probabilities can be approximated by the normal probability distribution? Explain. Yes, because $$ n p \geq 5 $$ and $$ n(1-p) \geq 5 $$, Yes, because $$ n p

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You may need to use the appropriate appendix table or technology to answer this question. Assume a binomial probability distribution has $$ p=0.60 $$ and $$ n=300 $$. (a) What are the mean and standard deviation? (Round your answers to two decimal places.) mean $$ \square $$ standard deviation $$ \square $$ (b) Is this situation one in which binomial probabilities can be approximated by the normal probability distribution? Explain. Yes, because $$ n p \geq 5 $$ and $$ n(1-p) \geq 5 $$, Yes, because $$ n p<5 $$ and $$ n(1-p)<5 $$, $$ \mathrm{No}_{	ext {, because }} n p \leqslant 5 $$ and $$ n(1-p) \leqslant 5_{1} $$ $$ \mathrm{No}_{1} $$ because $$ n p \geq 5 $$ and $$ n(1-p) \geq 5{ }_{1} $$ Yes, because $$ n \geqslant 30 $$. (c) What is the probability of 160 to 170 successes? Use the normal approximation of the binomial distribution to answer this question. (Round your answer to four decimal places.) $$ \square $$ (d) What is the probability of 190 or more successes? Use the normal approximation of the binomial distribution to answer this question. (Round your answer to four decimal places.) $$ \square $$ (e) What is the advantage of using the normal probability distribution to approximate the binomial probabilities? The advantage would be that using the normal probability distribution to approximate the binomial probabilities increases the number of calculations. The advantage would be that using the normal probability distribution to approximate the binomial probabilities reduces the number of calculations. The advantage would be that using the normal probability distribution to approximate the binomial probabilities makes the calculations more accurate. The advantage would be that using the normal probability distribution to approximate the binomial probabilities makes the calculations less accurate. How would you calculate the probability in part (d) using the binomial distribution. (Use $$ f(x) $$ to denote the binomial probability function.) $$ P(x \geq 190)=1-f(189)-f(190)-f(191)-f(192)-\ldots=f(300) $$ $$ P(x \geq 190)=f(0)+f(1)+\ldots+f(189)+f(190) $$ $$ P(x \geq 190)=f(191)+f(192)+f(193)+f(194)+\ldots+f(300) $$ $$ P(x \geq 190)=f(0)+f(1)+\ldots+f(188)+f(189) $$ $$ P(x \geq 190)=f(190)+f(191)+f(192)+f(193)+\ldots+f(300) $$

UpStudy AI · Accepted Answer

(a) The mean and standard deviation are calculated by

$$
\text{mean} = np = 300 \times 0.60 = 180
$$

$$
\text{standard deviation} = \sqrt{np(1-p)} = \sqrt{300 \times 0.60 \times 0.40} = \sqrt{72} \approx 8.49
$$

Thus, the mean is $$180$$ and the standard deviation is approximately $$8.49$$.

---

(b) The normal approximation is appropriate if both $$np$$ and $$n(1-p)$$ are at least 5. Here,

$$
np = 300 \times 0.60 = 180 \quad \text{and} \quad n(1-p) = 300 \times 0.40 = 120,
$$

both of which exceed 5. Therefore, the answer is:

Yes, because $$np \geq 5$$ and $$n(1-p) \geq 5$$.

---

(c) To approximate the probability of 160 to 170 successes using the normal distribution, we apply the continuity correction. We want

$$
P(160 \le X \le 170) \approx P(159.5 < X < 170.5)
$$

The $$z$$-scores are calculated as follows:

$$
z_1 = \frac{159.5 - 180}{8.49} \approx \frac{-20.5}{8.49} \approx -2.41
$$

$$
z_2 = \frac{170.5 - 180}{8.49} \approx \frac{-9.5}{8.49} \approx -1.12
$$

Using the standard normal distribution table:

$$
P(Z < -1.12) \approx 0.1314, \quad P(Z < -2.41) \approx 0.0080.
$$

Thus, the probability is

$$
P(159.5 < X < 170.5) = 0.1314 - 0.0080 = 0.1234.
$$

Rounded to four decimal places, the answer is $$0.1234$$.

---

(d) For 190 or more successes, we use the continuity correction $$X \geq 189.5$$:

$$
z = \frac{189.5 - 180}{8.49} \approx \frac{9.5}{8.49} \approx 1.12.
$$

Then,

$$
P(X \ge 189.5) = 1 - P(Z < 1.12).
$$

From the table, $$P(Z < 1.12) \approx 0.8686$$. Therefore,

$$
P(X \ge 189.5) = 1 - 0.8686 = 0.1314.
$$

Rounded to four decimal places, the probability is $$0.1314$$.

---

(e) The advantage of using the normal probability distribution to approximate the binomial probabilities is that it reduces the number of calculations.

To calculate the probability in part (d) using the binomial distribution, we would need to sum the binomial probabilities:

$$
P(x \geq 190) = f(190) + f(191) + f(192) + \ldots + f(300)
$$

Thus, the correct expression is

$$
P(x \geq 190)=f(190)+f(191)+f(192)+\ldots+f(300).
$$
(a) Mean: 180, Standard Deviation: 8.49 (b) Yes, because $$ np \geq 5 $$ and $$ n(1-p) \geq 5 $$ (c) Probability: 0.1234 (d) Probability: 0.1314 (e) Advantage: Using the normal distribution reduces the number of calculations. For part (d), the probability is calculated as $$ P(x \geq 190) = f(190) + f(191) + f(192) + \ldots + f(300) $$.

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