please answer these questions-thank you Question #1: The amount of fill (weight

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please answer these questions-thank you

Question #1: The amount of fill (weight in ounces) put into a glass jar of spaghetti sauce is normally distributed with a mean of =850 grams and =8 grams. The words "describe the distribution" mean to describe or compute the shape, mean, and spread of the distribution. a) Find the probability that one jar selected at random contains between 848 and 855 grams. Also, give the shape, center, and spread of the model you used to answer this question. b) Suppose that a sample of 25 jars of sauce have been selected. Find the probability that a random sample of 25 jars has a mean weight between 848 and 855 grams. Also, give the shape, center, and spread of the model you used to answer this question. c) Why is the answer to part (b) much more than the answer to part (a)? (Hint, draw a picture!)

Question #2: The local bakery bakes more than a thousand 1-pound loaves of bread daily, and the weights of these loaves varies. The mean weight is 1 lb and 1 oz, or 482 grams. Assume the standard deviation of the weights is 18 grams and a sample of 45 loaves is to be randomly selected. a) What is the shape, center, and spread of the model that you'll use to answer this question? b) What is the probability that this sample mean will be between 475 and 495 grams? c) What is the probability that the sample mean will have a value less than 478 grams? d) What is the probability that the sample mean will be within 5 grams of the mean?

Question #3: According to recent figures, 6% of college graduates (aged 21-24) are unemployed, which is the lowest rate since before the recession. Suppose we take a random sample of 230 recent graduates (aged 21-24). What is the probability that the unemployment rate is 9% or higher? (This 9% number is significant because it was the unemployment rate for college grads during the recession.) Does it seem like the unemploymen

Explanation / Answer

#Q1.

the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 850
standard Deviation ( sd )= 8
a.
BETWEEN THEM
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 848) = (848-850)/8
= -2/8 = -0.25
= P ( Z <-0.25) From Standard Normal Table
= 0.4013
P(X < 855) = (855-850)/8
= 5/8 = 0.625
= P ( Z <0.625) From Standard Normal Table
= 0.734
P(848 < X < 855) = 0.734-0.4013 = 0.3327
b.
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
mean of the sampling distribution ( x ) = 850
standard Deviation ( sd )= 8/ Sqrt ( 25 ) =1.6
sample size (n) = 25
BETWEEN THEM
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 848) = (848-850)/8/ Sqrt ( 25 )
= -2/1.6
= -1.25
= P ( Z <-1.25) From Standard Normal Table
= 0.10565
P(X < 855) = (855-850)/8/ Sqrt ( 25 )
= 5/1.6 = 3.125
= P ( Z <3.125) From Standard Normal Table
= 0.99911
P(848 < X < 855) = 0.99911-0.10565 = 0.89346
c.
since the standard error value is very minimal for larger size
distribution and this result in larger proprtion rate of values are
reside around the mean value

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