Answer:
There is a 95% confidence that the sample has a mean between 158.92 pounds and 171.48 pounds
Step-by-step explanation:
Given that mean (μ) = 165.2 pounds, standard deviation (σ) = 12.4 pounds, sample size (n) = 15 crates. Confidence (C) = 95% = 0.95
α = 1 - C = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025
The z score of α/2 corresponds to the z score of 0.475 (0.5 - 0.025) which is 1.96. 
The margin of error (E) is given by:
The confidence interval = μ ± E = 165.2 ± 6.28 = (158.92, 171.48)
The confidence interval is between 158.92 pounds and 171.48 pounds. There is a 95% confidence that the sample has a mean between 158.92 pounds and 171.48 pounds
Answer:
And tha's equivalent to use this formula:
Step-by-step explanation:
For this case the 95%of the values are between the following two values:
(75 , 163)
And for this case we know that the variable of interest X "length of a movie" follows a normal distribution:
We can estimate the true mean with the following formula:
Now we know that in the normal standard distribution we know that we have 95% of the values between 1.96 deviations from the mean. We can find the value of the deviation with this formula:
And tha's equivalent to use this formula:
Answer:YES
Step-by-step explanation:
Answer:
4/5 or 0.8
Step-by-step explanation:
1/3x = 5/12
<em>cross</em><em> </em><em>multiply</em>
(12×1) = (5×3x)
<em>divide</em><em> </em><em>through</em> <em>by</em> <em>1</em><em>5</em>
x = 12/15
x = 4/5 <em>o</em><em>r</em><em> </em>0.8
There are two other fine ways to write 11/5 .
Ten of those fifths make 2 wholes, so 11/5 can be written as<u> 2 and 1/5</u> .
Or you could do the division that the fraction shows.
Divide 11 by 5, and find the other way to write it . . . . . <u>2.2</u> .
