Answer:
The 95% confidence interval for the fraction of all shoppers during the year whose visit was because of a coupon they'd received in the mail is (0.2016, 0.2694).
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of 
, and a confidence level of 
, we have the following confidence interval of proportions.
In which
z is the z-score that has a p-value of 
.
A store randomly samples 603 shoppers over the course of a year and finds that 142 of them made their visit because of a coupon they'd received in the mail.
This means that 
95% confidence level
So 
, z is the value of Z that has a p-value of 
, so 
.
The lower limit of this interval is:
The upper limit of this interval is:
The 95% confidence interval for the fraction of all shoppers during the year whose visit was because of a coupon they'd received in the mail is (0.2016, 0.2694).
X= 20 y= 100
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Answer:
8
Step-by-step explanation:
o989
Algebra is really easy when you try your hardest. What seems to be troubling you friend?
Answer:
The heaviest 5% of fruits weigh more than 747.81 grams.
Step-by-step explanation:
We are given that a particular fruit's weights are normally distributed, with a mean of 733 grams and a standard deviation of 9 grams.
Let X = <u><em>weights of the fruits</em></u>
The z-score probability distribution for the normal distribution is given by;
Z = 
~ N(0,1)
where, 
= population mean weight = 733 grams
= standard deviation = 9 grams
Now, we have to find that heaviest 5% of fruits weigh more than how many grams, that means;
P(X > x) = 0.05 {where x is the required weight}
P( > 
) = 0.05
P(Z > ) = 0.05
In the z table the critical value of z that represents the top 5% of the area is given as 1.645, that means;
x = 747.81 grams
Hence, the heaviest 5% of fruits weigh more than 747.81 grams.
