-2x+1=0
=0
x-1=0
x=1
Scarlett has x dimes and y quarters, having no less than 16 coins worth at
1 answer:
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Answer:
95% confidence interval for the average mpg of all minivans in the company's inventory is [15.22 mpg , 15.98 mpg].
Step-by-step explanation:
We are given that a company checks the miles per gallon (mpg) for a simple random sample of 100 minivans drawn from its current inventory.
The average mpg of these 100 minivans is 15.6 with a standard deviation of 1.9 mpg.
Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;
P.Q. = ~
where, = sample average mpg = 15.6 mpg
s = sample standard deviation = 1.9 mpg
n = sample of minivans = 100
= population average mpg
<em>Here for constructing 95% confidence interval we have used One-sample t test statistics because we don't know about population standard deviation.</em>
So, 95% confidence interval for the population mean, is ;
P(-1.987 < < 1.987) = 0.95 {As the critical value of t at 99 degree
of freedom are -1.987 & 1.987 with P = 2.5%}
P(-1.987 < < 1.987) = 0.95
P( <
<
) = 0.95
P( <
<
) = 0.95
<u>95% confidence interval for</u> = [
,
]
= [ ,
]
= [15.22 mpg , 15.98 mpg]
Therefore, 95% confidence interval for the average mpg of all minivans in the company's inventory is [15.22 mpg , 15.98 mpg].
It is appropriate to compute a confidence interval for this problem using the Normal curve as t test statistics is used when data should follow normal distribution.