In a store chain with separate bakery departments it is required, that the breads in all the chain’s bakery shops are reasonably homogeneous. For the weight of a given bread the following requirements are specified:
1% of the breads are allowed to weigh below 600 g.
5% of the breads are allowed to weigh above 650 g.
Data are assumed to follow a normal distribution
The mean and standard deviation $(\mu ,\sigma )$ to be used for the production to meet these requirements are:
First we find the relevant percentiles of the standard normal (Z) distribution:
\(P(Z<z_{0.99})=0.01 \mbox{ and } P(Z>z_{0.05})=0.05\)
Hence these values can be found from R as qnorm(0.01) and qnorm(0.95)
\(z_{0.99}=-2.326 \mbox{ and } z_{0.05}=1.645\)
We need to find the $\mu$ and the $\sigma$ that will give the same probabilities in a non-standard normal $X$, so we can use the basic standardization formula $Z=\frac{X-\mu}{\sigma}$ twice and solve 2 equations with two unknowns:
\(-2.326=\frac{x_{0.99}-\mu}{\sigma} \mbox { and } 1.645=\frac{x_{0.05}-\mu}{\sigma}\)
that is:
\(-2.326=\frac{600-\mu}{\sigma} \mbox { and } 1.645=\frac{650-\mu}{\sigma}\)
The solutions is the one given as:
$\mu = 629.3 and $\sigma =12.6$
Question 2 of 2
The following data are measurements of albumin in blood samples from a group of people consisting of 7 males and 8 females. The standard deviation and the mean of the measurements are given separately for the two groups:
Mean
Standard deviation
Males
38
36
37
41
37
43
41
$\overline{x}_1=39.0$
$s_1=2.6458$
Females
45
39
41
44
47
46
44
42
$\overline{x}_2=43.5$
$s_2=2.6726$
Assume that the female population, which the sample of the 8 females represents, consists of 100000 females.
Based on the estimated distribution for the females, how many females in the population are expected to have an albumin content in the blood equal to 48 or larger?
The mean and standard deviation of albumin content in women are given as 43.5 and 2.6726, respectively. Let $X$ denote the albumin in a single randomly chosen woman. Then $X \sim N(43.5, 2.6726)$. We find
\(P(X > 48) = 1 - P(X \leq 48)\)
In R, this can be found using pnorm as follows
