Question

The amount of fill (weight of contents) put into a glass jar of spaghetti sauce is normally distributed with mean μ = 840 grams and standard deviation of σ = 9 grams.

(a) Describe the distribution of x, the amount of fill per jar.

skewed right normal     skewed left chi-square

(b) Find the probability that one jar selected at random contains between 843 and 858 grams. (Give your answer correct to four decimal places.)

(c) Describe the distribution of x, the mean weight for a sample of 20 such jars of sauce.

skewed right normal     skewed left chi-square

(d) Find the mean of the x distribution. (Give your answer correct to the nearest whole number.)


(ii) Find the standard error of the x distribution. (Give your answer correct to two decimal places.)


(e) Find the probability that a random sample of 20 jars has a mean weight between 843 and 858 grams. (Give your answer correct to four decimal places.)

Answer 1

a)

Normal

b)

Here, μ = 840, σ = 9, x1 = 843 and x2 = 858. We need to compute P(843<= X <= 858). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (843 - 840)/9 = 0.33
z2 = (858 - 840)/9 = 2

Therefore, we get
P(843 <= X <= 858) = P((858 - 840)/9) <= z <= (858 - 840)/9)
= P(0.33 <= z <= 2) = P(z <= 2) - P(z <= 0.33)
= 0.9772 - 0.6293
= 0.3479

c)

normal

d)

mena = 840

2)

standard error = 9/sqrt(20) = 2.01

e)

Here, μ = 840, σ = 2.01, x1 = 843 and x2 = 858. We need to compute P(843<= X <= 858). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (843 - 840)/2.01 = 1.49
z2 = (858 - 840)/2.01 = 8.96

Therefore, we get
P(843 <= X <= 858) = P((858 - 840)/2.01) <= z <= (858 - 840)/2.01)
= P(1.49 <= z <= 8.96) = P(z <= 8.96) - P(z <= 1.49)
= 1 - 0.9319
= 0.0681