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Question 6 2 pts

In an agricultural study, the average amount of corn yield is normally distributed with a mean of 189.3 bushels of corn per acre, with a standard deviation of 23.5 bushels of corn. If a study included 1200 acres, about how many would be expected to yield more than 180 bushels of corn per acre?

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Question 7 2 pts

On average, the parts from a supplier have a mean of 31.8 inches and a standard deviation of 2.4 inches. Find the probability that a randomly selected part from this supplier will have a value between 29.4 and 34.2 inches. Is this consistent with the Empirical Rule of 68%-95%-99.7%?

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Question 8 2 pts

A process is normally distributed with a mean of 10.6 hits per minute and a standard deviation of 0.49 hits. If a randomly selected minute has 11.8 hits, would the process be considered in control or out of control?

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Question 9 2 pts

The candy produced by a company has a sugar level that is normally distributed with a mean of 16.1 grams and a standard deviation of 0.9 grams. The company takes readings of every 10th bar off the production line. The reading points are 17.3, 14.9, 18.3, 16.5, 16.1, 17.4, 19.4. Is the process in control or out of control and why?

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Question 10 2 pts

The toasters produced by a company have a normally distributed life span with a mean of 5.8 years and a standard deviation of 0.9 years, what warranty should be provided so that the company is replacing at most 4% of their toasters sold?

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 ) = 189.3
standard Deviation ( sd )= 23.5
GREATER THAN
P(X > 180) = (180-189.3)/23.5
= -9.3/23.5 = -0.3957
= P ( Z >-0.3957) From Standard Normal Table
= 0.6539
study included 1200 acres, no.of expected to yield more than 180 bushels of corn per acre = 1200 * 0.6539
= 785 acres

Q2.
About 68% of the area under the normal curve is within one standard deviation of the mean. i.e. (u ± 1s.d)
So to the given normal distribution about 68% of the observations lie in between
= [31.8 ± 2.4]
= [ 31.8 - 2.4 , 31.8 + 2.4]
= [ 29.4 , 34.2 ]

Probability is 0.68, which is consistent with the Empirical Rule

Q3.
P(X > 11.8) = (11.8-10.6)/0.49
= 1.2/0.49 = 2.449
= P ( Z >2.449) From Standard Normal Table
= 0.0072 > 5% , can be considered to in control

In control as this one data point is not more than three standard deviations from the mean

Q5.
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 ) = 5.8
standard Deviation ( sd )= 0.9
UPPER/TOP
P ( Z > x ) = 0.04
Value of z to the cumulative probability of 0.04 from normal table is 1.750686
P( x-u / (s.d) > x - 5.8/0.9) = 0.04
That is, ( x - 5.8/0.9) = 1.750686
--> x = 1.750686 * 0.9+5.8 = 7.375617 ~ 7.3 years

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