Complete the following question using Excel. A local business knows that 30% of
ID: 3365927 • Letter: C
Question
Complete the following question using Excel.
A local business knows that 30% of their customers use a credit card to pay for their purchase. Suppose a sample of 85 customers is selected.
What is the distribution of the sample proportion? Enter "binomial" or "normal" in blank #1.
In blank #2 enter the probability that less than 18 customers in the sample use a credit card to pay for their purchases. (Round your answer to 4 decimal places and enter your answer as 0.1234, not .1234.)
In blank #3 enter the probability that more than 35% of customers in the sample use a credit card to pay for their purchases. (Round your answer to 4 decimal places and enter your answer as 0.1234, not .1234.)
In blank #4 enter the probability that between 20 and 30 customers in the sample use a credit card to pay for their purchases. (Round your answer to 4 decimal places and enter your answer as 0.1234, not .1234.)
Explanation / Answer
PART A.
Points to pass for normal approximation:
1) experiment consistes of a sequence of n identical trials
2) only 2 outcomes are possible on each trail, success or failure
3) trials are independent & below conditions should satisfy
n*p>5, 85*0.3> 5 => 25.5>5
n*(1-p)>5, 85*0.7> 5 => 59.5>5
can use normal approximation
It is normal
PART B.
P(X < 18) = (18-25.5)/4.2249
= -7.5/4.2249= -1.7752
= P ( Z <-1.7752) From Standard NOrmal Table
= 0.0379
PART C.
P(X > 0.35) = (0.35-0.3)/0.0497
= 0.05/0.0497 = 1.006
= P ( Z >1.006) From Standard Normal Table
= 0.1572
PART D.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 20) = (20-25.5)/4.2249
= -5.5/4.2249 = -1.3018
= P ( Z <-1.3018) From Standard Normal Table
= 0.09649
P(X < 30) = (30-25.5)/4.2249
= 4.5/4.2249 = 1.0651
= P ( Z <1.0651) From Standard Normal Table
= 0.85659
P(20 < X < 30) = 0.85659-0.09649 = 0.7601