Answer:
Solution given
:total n[U]=590
customers using coupon n[C]=177
the percentage of all customers who often use coupons=177/590×100%=30%.
If n[U]=5000
no of customer using coupon be x
we have
30% of 5000=x
x=30/100×5000
x=1500
<h3>
<u>So1500 people</u><u> </u>would you expect to use a coupon if all 5,000 people in the community visited the store.</h3>
Answer:
Probability that the sample variance exceeds 3.10 is less than 0.05%.
Step-by-step explanation:
We are given that a process produces batches of a chemical whose impurity concentrations follow a normal distribution with a variance of 1.75.
Also, a random sample of 20 of these batches is chosen.
The probability distribution that we will used here is of Chi-square distribution, i.e.;
~
where, = sample variance
= population variance = 1.75
n = sample of batches = 20
So, probability that the sample variance exceeds 3.10 is given by = P( > 3.10)
P( > 3.10) = P( > )
= P( > 59.62) = P( > 59.62) = Less than 0.05%
From chi-square table we can observe that Probability that is greater than is less than 0.05%.
So, the probability that the sample variance exceeds 3.10 is less than 0.05%.
Answer:
something
Step-by-step explanation: