Question

The time of a manufacturer needs to produce a tonne of fertiliser is known to be normally distributed with a mean of 420 minutes and a variance of 900 minutes. The sample

(a) What is the probability of the time needed between 390 and 480 minutes? Sketch a probability distribution curve to show the probability value, 𝑍 value and its mean value.

(b) If the probability is 0.125, the time needed will be less than how many minutes? Sketch a probability distribution curve to show the time needed and the mean value.

(c) The manufacturer takes a sample of 20 tonnes of fertiliser. If the sample mean time needed is above 430 minutes, the process is deemed inefficient. What is the probability that the process is inefficient?

(d) Refer to (c) for sample size and mean. Find the upper and lower confidence limits of a 90% confidence interval for the population mean.

(e) Identify the point estimator and use the information in this question to explain small sample properties.

Answer 1

a) $z=\frac{480-420}{30}=2$; propability z=2 is 0.4772

$z=\frac{390-420}{30}=-1$; propability z=-1 is 0.1587

the probability of the time needed between 390 and 480 minutes:

$0.4772-0.1587=0.3185$

b) Probability -1.25 is z=-1.15

$-1.15=\frac{x-420}{30}$; $x=385.5$ minutes.

c)

$\sigma_{20}=\frac{30}{\sqrt{\smash[b]{20}}}=6.71$

$z=\frac{430-420}{6.71}=1.49$

Propability: 0.9319

d) Lower confidence limits

propability: 0.05 z=-1.645

$-1.645=\frac{x-420}{6.71}$; $x=408.96$

Upper confidence limits

propability: 0.95 z=1.645

$1.645=\frac{x-420}{6.71}$; $x=431.04$

confidence interval 409-431 minutes

e)  the point estimator:

$\frac{(2*20-1)*6.71^4}{20^2}=197.65$

197.65<420