<span>a.
</span>Do you
have sufficient funds to estimate the population mean for the attribute of
interest with a 95% confidence interval 4 units width? Assume that sd= 12
n= {[(Zalpha/2)^2]*[sd]^2}/
se^2
n=
(1.96)^2*(12)^2/ (2)^2
n=
138.297 rounded up to 139
<span>There
is not enough funds for this one
since you’ll need 139 pieces and it costs 10 a piece, you’ll need 1390.</span>
b.
90% confidence interval
n= {[(Zalpha/2)^2]*[sd]^2}/
se^2
n=
(1.645)^2*(12)^2/ (2)^2
n=98
There is enough
funds since 98 pieces for 10 a piece is equal to 980.
Answer:
$51,862.50
Step-by-step explanation:
... the product of the number of shares and the price:
(900)($57 5/8) = $51,862.50
Step-by-step explanation:
The solution to this problem is very much similar to your previous ones, already answered by Sqdancefan.
Given:
mean, mu = 3550 lbs (hope I read the first five correctly, and it's not a six)
standard deviation, sigma = 870 lbs
weights are normally distributed, and assume large samples.
Probability to be estimated between W1=2800 and W2=4500 lbs.
Solution:
We calculate Z-scores for each of the limits in order to estimate probabilities from tables.
For W1 (lower limit),
Z1=(W1-mu)/sigma = (2800 - 3550)/870 = -.862069
From tables, P(Z<Z1) = 0.194325
For W2 (upper limit):
Z2=(W2-mu)/sigma = (4500-3550)/879 = 1.091954
From tables, P(Z<Z2) = 0.862573
Therefore probability that weight is between W1 and W2 is
P( W1 < W < W2 )
= P(Z1 < Z < Z2)
= P(Z<Z2) - P(Z<Z1)
= 0.862573 - 0.194325
= 0.668248
= 0.67 (to the hundredth)
Answer:
F
Step-by-step explanation:
Area of sector=pi*r^2*(theta/360) = pi*r^2*(45/360)
18*pi=pi*r^2*(1/8), r=12. Circumference is 24*pi
