Question

One reason the Normal approximation may fail to give accurate estimates of binomial probabilities is that the binomial distributions are discrete and the Normal distributions are continuous. That is, counts take only whole number values, but Normal variables can take any value. We can improve the Normal approximation by treating each whole number count as if it occupied the interval from 0.50.5 below the number to 0.50.5 above the number. For example, approximate a binomial probability ?(?≥10)P(X≥10) by finding the Normal probability ?(?≥9.5)P(X≥9.5) . Be careful: binomial ?(?>10)P(X>10) is approximated by Normal ?(?≥10.5)P(X≥10.5) .

According to CDC estimates, more than 22 million people in the United States are sickened each year with antibiotic‑resistant infections, with at least 23,00023,000 dying as a result. Antibiotic resistance occurs when disease‑causing microbes become resistant to antibiotic drug therapy. Because this resistance is typically genetic and transferred to the next generations of microbes, it is a very serious public health problem. Of the three infections considered most serious by the CDC, gonorrhea has an estimated 800,000800,000 cases occurring annually, with approximately 30%30% of those cases resistant to any antibiotic.

Suppose a local health clinic sees 4040 cases. The exact binomial probability that 1515 or more cases are resistant to any antibiotic is 0.1930.193 .

(a) Show that this setting satisfies the rule of thumb for the use of the Normal approximation (just barely). (Enter your answers rounded to one decimal place.)

??=np=

?(1−?)=n(1−p)=

(b) What is the Normal approximation to ?(?≥15)P(X≥15) ? (Enter your answer to four decimal places.)

?(?≥15)=P(X≥15)=

It can be observed that the Normal approximation using continuity correction gives a much closer value to the true binomial probability than the Normal approximation.

(c) What is the Normal approximation using the continuity correction? (Enter your answer to four decimal places.)

?(?≥14.5)=P(X≥14.5)=

Answer 1

(a)

n = 40

p = 0.30

q = 1 - p = 0.70

np = 40 X 0.30 = 12 > 10

n (1 - p) = 40 X 0.7 = 28 > 10

Thus, this setting satisfies the rule of thumb for the use of the Normal approximation.

(b)

= np = 40 X 0.30 = 12

= sqrt(40 X0.3 X 0.7) = 2.8983

To find P(X15):

Z = (15 - 12)/2.8983 = 1.04

Table of Area Under Standard Normal Curve gives area = 0.3508

So,

P(X15) = 0.5 - 0.3508 = 0.1492

So,

Answer is:

0.1492

(c)

= np = 40 X 0.30 = 12

= sqrt(40X0.3X0.7) = 2.8983

To find P(X14.5):

Z = (14.5 - 12)/2.8983 = 0.86

Table of Area Under Standard Normal Curve gives area = 0.3051

So,

P(X14.5) = 0.5 - 0.3051 = 0.1949

So,

Answer is:

0.1949