Question

1)A new chemical has been found to be present in the human bloodstream, and a medical group would like to study the presence of this chemical in some samples of patients. The presence of the chemical in a patient is measured by a score representing the 'parts per billion' in which that chemical appears in the blood. It is known that, on this scale, men have an average score of 810.9 and a standard deviation of 58. It is also known that women have an average score of 835.48 and a standard deviation of 21.

An assistant in the medical team has been handed a sample of 100 scores. The assistant knows that all of the scores are from one of the two genders, but the sample was not documented very well and so they do not which gender this is. Within the sample, the mean score is 825.4.

a)Complete the following statements. Give your answers to 1 decimal place.

If the sample came from a group of 100 men, then the sample mean is ______ standard deviations above the mean of the sampling distribution. In contrast, if the sample came from a group of 100 women, then the sample mean is _______ standard deviations below the mean of the sampling distribution.

b)Based on this, the assistant is more confident that the sample came from a group of 100 _____men or women_____

2)The life span at the birth of humans has a mean of 87.74 years and a standard deviation of 17.76 years. Calculate the upper and lower bounds of an interval containing 95% of the sample mean life spans at birth based on samples of 105 people. Give your answers to 2 decimal places.

a)Upper bound = _________ years

b)Lower bound = ______ years

3)A drug made by a pharmaceutical company comes in tablet form. Each tablet is branded as containing 120 mg of the particular active chemical. However, variation in manufacturing results in the actual amount of the active chemical in each tablet following a normal distribution with mean 120 mg and standard deviation 1.665 mg.

a)Calculate the percentage of tablets that will contain less than 119 mg of the active chemical. Give your answer as a percentage to 2 decimal places.

Percentage = %

b)Suppose samples of 12 randomly selected tablets are taken and the amount of active chemical measured. Calculate the percentage of samples that will have a sample mean of less than 119 mg of the active chemical. Give your answer as a percentage to 2 decimal places.

Percentage = %

4)

During its manufacturing process, Fantra fills its 20 fl oz bottles using an automated filling machine. This machine is not perfect and will not always fill each bottle with exactly 20 fl oz of soft drink. The amount of soft drink poured into each bottle follows a normal distribution with mean 20 fl oz and a standard deviation of 0.17 fl oz.

The Fantra quality testing department has just carried out a routine check on the average amount of soft drink poured into each bottle. A sample of 25 bottles was randomly selected and the amount of soft drink in each bottle was measured. The mean amount of soft drink in each bottle was calculated to be 19.90 fl oz. The Fantra Chief Executive Officer believes that such a low mean is not possible and a mistake must have been made.

Calculate the probability of obtaining a sample mean below 19.90 fl oz. Give your answer as a decimal to 4 decimal places.

probability =

Answer 1

a) z-score for men = ()/()

                              = (825.4 - 810.9)/(58/)

                              = 2.5

z-score for women = ()/()

                              = (825.4 - 835.48)/(21/)

                              = -4.8

If the sample came from a group of 100 men , then the sample mean is 2.5 standard deviations above the mean of the sampling distribution. In contrast if the sample came from a group of 100 women , then the sample mean is 4.8 standard deviations below the mean of the sampling distribution.

b) Based on this, , the assistant is more confident that the sample came from a group of 100 men.

2) At 95% confidence interval the critical value is z_0.025 = 1.96

The 95% confidence interval is

+/- z_0.025 *

= 87.74 +/- 1.96 * 17.76/

= 87.74 +/- 3.397

= 84.343, 91.137

= 84.34, 91.14

a) Upper bound = 91.14

b) Lower bound = 84.34

3) a) P(X < 119)

= P((X - )/ < (119 - )/)

= P(Z < (119 - 120)/1.665)

= P(Z < -0.60)

= 0.2743 = 27.43%

b) P( < 119)

= P(( - )/() < (119 - )/())

= P(Z < (119 - 120)/(1.665/))

= P(Z < -2.08)

= 0.0188 = 1.88%

4) P( < 19.9)

= P(( - )/() < (19.9 - )/())

= P(Z < (19.9 - 20)/(0.17/))

= P(Z < -2.94)

= 0.0016