Question

EcoClear Products manufactures RatX, a 100% non-toxic one-of-a-kind natural humane method to kill rats and mice that is 100% safe for people, pets and all other wildlife. Quality control personnel at EcoClear Products have determined that Bags of RatX have a population mean of 16 ounces and a population standard deviation of 4 ounces.

Part 1

A random sample of 64 bags of RatX is drawn. The probability that the average weight of the bags of RatX will be more than 17.26 ounces is . Use only the appropriate formula and/or statistical table in your textbook to answer this question. Report your answer to 4 decimal places, using conventional rounding rules.

Part 2

A random sample of 64 bags of RatX is drawn. The probability that the average weight of the bags of RatX will be between 15.36 and 15.68 ounces is . Use only the appropriate formula and/or statistical table in your textbook to answer this question. Report your answer to 4 decimal places, using conventional rounding rules.

Part 3

A random sample of 64 bags of RatX is drawn. The probability that the average weight of the bags of RatX will be less than 17.21 ounces is . Use only the appropriate formula and/or statistical table in your textbook to answer this question. Report your answer to 4 decimal places, using conventional rounding rules.

Part 4

Assume that random samples of 64 bags of RatX have been repeatedly drawn from the population and the mean number of ounces in each sample is calculated. Eighty-six percent (86%) of the sample means should be above ounces. Use only the appropriate formula and/or statistical table in your textbook to answer this question. Report your answer to 2 decimal places, using conventional rounding rules.

Answer 1

Solution :

Given = 16 ounces and = 4 ounces

(1)

n = 64

For P (X > 17.6) = 1 - P (X < 17..6), as the normal tables give us the left tailed probability only.

For P( X < 17.6)

Z = (17.6 – 16)/[ 4 / sqrt(64)] = 3.2

The probability for P(X < 17.6) from the normal distribution tables is = 0.9993

Therefore the required probability = 1 – 0.9993 = 0.0007

(2)

Between 15.36 and 15.68 = P(15.68 < X < 15.36) = P(X < 15.68) - P(X < 15.36)

For P(X < 15.68) ; z = (15.68 - 16) / [4 / sqrt(64]) = -0.64. The p value at this score is = 0.2611

For P(X < 15.36) ; z = (15.36 - 16) / [4 / sqrt(64]) = -1.28. The p value at this score is = 0.1003

Therefore the required probability is 0.2611 – 0.1003 = 0.1608

(3)

n = 64

For P (X < 17.21)

Z = (17.21 – 16)/[ 4 / sqrt(64)] = 3.2

The probability for P(X < 17.21) from the normal distribution tables is = 0.9922

(4)

if 86% should be above this value, then 14% will be below this

P(X > x) 0.14

The z score at p = 0.14 is -1.08

Therefore -1.08 = (X - 16) / [4 / sqrt(64)

Solving for X, we get X = (-1.08 * 4 / 8) + 16

X = 15.46

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