To assess the accuracy of a laboratory scale, a standard weight known to weigh 10 grams is weighed repeatedly. The scale readings are Normally distributed with unknown mean (this mean is 10 grams if the scale has no bias). The standard deviation of the scale readings is known to be 0.0002 gram.

a. The weight is measured five times. The mean result is 10.0023 grams. Give a 98% confidence interval for the mean of repeated measurements of the weight.
b. How many measurements must be averaged to get a margin of error of 20.0001 with 98% confidence?

Step-by-step explanation:

a) Confidence interval is written in the form,

(Sample mean - margin of error, sample mean + margin of error)

The sample mean, x is the point estimate for the population mean.

Since the population standard deviation is known, we would determine the z score from the normal distribution table.

Margin of error = z × σ/√n

Where

σ = population standard Deviation

n = number of samples

From the information given

x = 10.0023 grams

σ = 0.0002 gram

n = 5

To determine the z score, we subtract the confidence level from 100% to get α

α = 1 - 0.98 = 0.02

α/2 = 0.02/2 = 0.01

This is the area in each tail. Since we want the area in the middle, it becomes

1 - 0.01 = 0.99

The z score corresponding to the area on the z table is 2.326. Thus, confidence level of 98% is 2.326

Margin of error = 2.326 × 0.0002/√5 = 0.00021

Confidence interval = 10.0023 ± 0.00021

b) Margin of error = ±0.0001

z = 2.326

Therefore

0.0001 = 2.326 × 0.0002/√n

0.0001/2.326 = 0.0002/√n

0.00004299226 = 0.0002/√n

√n = 0.0002/0.00004299226 = 4.65

n = 4.65² = 21.6

n = 22

45 grams

Step-by-step explanation:

1st order: 27 people

15 people:

100 g

Ginger: 20 g

Turmeric: 25 g

Coriander: 35 g

Cumin Seeds: 20 g

20+25+35+20=60+40=100

For 1 person:

Ginger: 1 1/3 g

Turmeric: 1 2/3 g

Coriander: 2 1/3 g

Cumin Seeds: 1 1/3 g

Multiply all that by 27

For 27 people:

Ginger: 36 g

Turmeric: 45 g

Coriander: 63 g

Cumin Seeds: 36 g

45 g for Turmeric