Question

1. A certain medical test is known to detect 72% of the people who are afflicted with the disease Y. If 10 people with the disease are administered the test, what is the probability that the test will show that: All 10 have the disease, rounded to four decimal places? .0374 At least 8 have the disease, rounded to four decimal places? At most 4 have the disease, rounded to four decimal places? Please show the steps in Microsoft excel 2016 if possible! Thank you

2.

According to government data, 24% of employed women have never been married. Rounding to 4 decimal places, if 15 employed women are randomly selected:

a. What is the probability that exactly 2 of them have never been married?

b. That at most 2 of them have never been married?

c. That at least 13 of them have been married

Please show work in excel

3.

Assume that 83% of people are left-handed. If we select 5 people at random, find the probability of each outcome described below, rounded to four decimal places:

a. There are some lefties (≥ 1) among the 5 people.

b. There are exactly 3 lefties in the group.

c. There are at least 4 lefties in the group.

d. There are no more than 2 lefties in the group.

e. How many lefties do you expect?

f. With what standard deviation?

please show work in excel

Answer 1

1) Given

p = 0.72 , q= 1-0.72 = 0.28 , n= 10

Probability that all 10 have this disease

By using Excel

P(x=10) = BINOMDIST(10,10,0.72,0) = 0.0374

Probability of at least 8 have the disease

P(x >=8) = 1 -  BINOMDIST(7,10,0.72,1)

= 1- 0.5622

= 0.4378

Probability of at most 4 have the disease

P(x<=4) = BINOMDIST(4,10,0.72,1) = 0.0342

According to HomeworkLib we have to answer first question.