a)
µ = 200
σ = 16.30
we need to calculate probability for ,
183.7 ≤ X ≤ 216.3
X1 = 183.7 , X2 =
216.3
Z1 = (X1 - µ ) / σ = -1.000
Z2 = (X2 - µ ) / σ = 1.000
P ( 183.7 < X <
216.3 ) = P ( -1
< Z < 1.000 )
= P ( Z < 1.000 ) - P ( Z
< -1.000 ) =
0.8413 - 0.1587 =
0.6827
b)
µ = 200
σ = 16.30
right tailed
X ≥ 216.3
Z = (X - µ ) / σ = 1.00
P(X ≥ 216.3 ) = P(Z ≥
1.00 ) = P ( Z <
-1.00 ) = 0.1587
c)
µ = 200
σ = 16.30
we need to calculate probability for ,
178 ≤ X ≤ 225
X1 = 178 , X2 =
225
Z1 = (X1 - µ ) / σ = -1.350
Z2 = (X2 - µ ) / σ = 1.534
P ( 178 < X <
225 ) = P (
-1.349693252 < Z < 1.534
)
= P ( Z < 1.534 ) - P ( Z
< -1.350 ) =
0.9375 - 0.0886 =
0.8489
so, 0.8489 proportion of boxes contain between 178g and 225 g
so, total numbe r of boxes = 200*0.8489 = 169.779 (so, approximately 170 boxes)
d)
µ = 200
σ = 16.3
proportion= 0.90
so, we will calculate Z value for 0.05 , and 0.95
Z value at 0.05 =
-1.645 (excel formula =NORMSINV(α))
z=(x-µ)/σ
so, X=zσ+µ= -1.645 *
16.3 + 200
X = 173.1889 g
Z value at 0.95 =
1.645 (excel formula =NORMSINV(α))
z=(x-µ)/σ
so, X=zσ+µ= 1.645 *
16.3 + 200
X = 226.8111g
so, answer is (173.19g, 226.81g)
Show steps 53.A manufacturer of cereal finds that the masses of cereal in the company's 200-g...
The answer for b) 0.159 c) 0.849
d) 200g +- 26.8g
Show steps
53. A manufacturer of cereal finds that the masses of cereal in the company's 200-g packages are normally distributed with a mean of 200 g and a standard deviation of 16.3 g. a) What proportion of these boxes contain between 183.7 g and 216.3 g of cereal? selected at random contains more than 216.3 g of cereal? would you expect to contain between 178 g and 225...
Boxes are labeled as containing 400 g of cereal. The machine filling the boxes produces weights that are normally distributed with standard deviation 12 g. (a) If the target weight is 400 g, what is the probability that the machine produces a box with less than 375 g of cereal? (Round your answer to three decimal places.) (b) Suppose a law states that no more than 10% of a manufacturer's cereal boxes can contain less than the stated weight of...
The amount of cereal that can be poured into a small bowl is normally distributed with a mean of 1.5 ounces and a standard deviation of 0.3 ounces. For a large bowl the amount poured is normally distributed a mean of 2.5 ounces with a standard deviation of 0.4 ounces. You open a new box of cereal and pour one large and one small bowl. Let Y be the difference of the amount of cereal in the two bowls (Large-S...
hello i need help solving this problems please
c. In a group of 230 tests, how many students score within one standard deviation of the men? 19. The number of roils of a given length is normally distributed with a mean length of 5.00 in and a standard deviation of 0.03 in a. Find the number of nails in a bag of 120 thet are less than 494 in long b. Find the number of nails in a bag of...
of contldence, the 00 of the 200 UCLA university professors oppose the US policies in Middle East. Using the 0.95 degree interval estimates for the population proportion is a, 50% to 57%. 45% to 55% d. none of the above. A Rutgers University professor claims that there is no significant difference between proportion of female (1) and male collects data students (m) of the university who exercise at least 15 minutes a day. What would be a proper conclusion if...
Question 202.5 pts If we consider the simple random sampling process as an experiment, the sample mean is _____. Group of answer choices always zero known in advance a random variable exactly equal to the population mean Flag this Question Question 212.5 pts The basis for using a normal probability distribution to approximate the sampling distribution of x ¯ and p ¯ is called _____. Group of answer choices The Law of Repeated Sampling The Central Limit Theorem Expected Value...