Question

# 23 (CONtinwed (C) Find the probability that fewer than 95 workers wil say ye (Round to four decimal places as needed ) Sketch the graph of the normal distribution with the indicated probability shaded O A. ○B. O c. O D. The normal distribution cannot be used to approximate the binomial distribution μ-897G 89 78 ,.8976 (d) Identify any unusual events. Explain □ A. There are no unusual events, because all of the probabilites are greater than 0。5 □ B. The event in part (c) is unusual because its probablity is less than 0 05 □ c. The event in part (afis unusual because its probabilty is less than 0.05. □ D. The event in part (b) is unusual because its probablity aless than 0 05. 24. Decide whether you can use the normal disinbution to approximate the binomial distribution. If you can, use the normal distribution to approximate the indicated probablities and sketch their graphs. If you cannot, explain why and use the binomial distribution to find the indicated probabilities. Five percent of workers in a city use public transportation to get to work. You randomly select 257 workers and ask them i they use public transportation to get to work. Complete parts (a) through (d) Can the normal distribution be used to approximate the binomial distribution? A. No, because ng<5. O C. No, because np 5 (a) Find the probability that exactly 18 workers will say yes What is the indicated probability? B. Yes because both np 2 5 and nq2 5 (Round to four decimal places as needed.) Sketch the graph of the normal distribution with the indicated probability shaded O A. O B. O c. The normal distribution cannot be used to approximate the binomial distribution. μ 1285 12 85 1285 付13

Answer 1

From the provided information we have,

= probability that workers use public transportation = 0.05 (5 percent)
= number of workers selected = 257

If a binomial distribution satisfied the requirements that
np
and $nq\geq 5$ then the binomial probability distribution can be approximated by a normal distribution with mean, $\mu =np$ and standard deviation, $\sigma =\sqrt{np(1-p)}$ and with the discrete whole number adjusted with a continuity correction, so that is represented by the interval and
X 0.5
.

So the normal distribution can be used as approximation though there will be a small error as is small and close to 0 (the approximation works best for values of around 0.5).
For this example, we have
$np=0.05*257=12.85\geq 5$

$nq=257*0.95=244.15\geq 5$

Yes, Normal Distribution can be used.

a)

We have,

$\mu =np=12.85$ and

$\sigma =\sqrt{np(1-p)}$

$\sigma =\sqrt{12.85*0.95}$

$\sigma =3.494$

Thus,

$X\sim B(257,0.05)\Rightarrow X\sim N(12.85,3.494)$

To find the probability that exactly 18 workers will say yes we need to find the corresponding Z-score using the following formula,

$Z=(X-\mu )/\sigma$

Z = (18 – 12.85) / 3.494

Z = 1.474

Therefore,

P[ X= 18 workers use public transportation] = P(Z = 1.474)

                                                                      = 0.1405

The graph of the normal distribution with the indicated probability is given by,

J3 18