Question

1) The vegetables grown in a farm have a mean weight of 23.2 ounces with a standard deviation of 1.5 ounces. Assuming that the distribution of the weights of these vegetables has roughly the shape of a normal distribution, what percentage of the vegetable weight

a) Less than 17.1 ounces

b) More than 19.3 ounces

c) Anywhere from 18.7 to 21.3 ounces

2)In a pollution study of the air in a certain downtown area, an EPA technician obtained a mean of 4.55 microgram of suspended bezene soluble matter per cubic meter with a standart deviation of 0.72 microgram for a sample of size n=9. Assuming that the population sampled is normal,

a) Construct a 90% confidence interval for the mean of the population sampled;

b) What can the technician assert with 95% confidence about the maximum error if x (mean)=4.55 micrograms per cubic meter is used as an estimate of the mean of the population sampled? 

Answer 1

1) Let $X=$ the vegetable weight: $X\sim N(\mu, \sigma^2)$

Given $\mu=23.2\ ounces, \sigma=1.5\ ounces$

a)

$$P(X<17.1)=P(Z<\dfrac{17.1-23.2}{1.5})$$

$$\approx P(Z<-4.0666666667)\approx0.0000238452$$

$0.0024\%$

b)

$$P(X>19.3)=1-P(Z\leq\dfrac{19.3-23.2}{1.5})$$

$$=1- P(Z\leq-2.6)\approx0.995339$$

$99.5339\%$

c)

$$P(18.7<X<21.3)=P(X<21.3)-P(X\leq18.7)$$

$$=P(Z<\dfrac{21.3-23.2}{1.5})-P(Z\leq\dfrac{18.7-23.2}{1.5})$$

$$\approx P(Z<-1.266667)-P(Z\leq-3)$$

$$\approx0.10263725-0.00134990\approx0.10128735$$

$10.1287\%$

2) The critical value for $\alpha=0.1$ and $df=n-1=8$ degrees of freedom is $t_c=1.859547.$

a0 The corresponding confidence interval is computed as shown below:

$$CI=(\bar{x}-t_c\times\dfrac{s}{\sqrt{n}}, \bar{x}+t_c\times\dfrac{s}{\sqrt{n}})$$

$$=(4.55-1.859547\times\dfrac{0.72}{\sqrt{9}},$$

$$4.55+1.859547\times\dfrac{0.72}{\sqrt{9}})$$

$$=(4.1037, 4.9963)$$

Therefore, based on the data provided, the 90% confidence interval for the population mean is $4.1037<\mu<4.9963,$ which indicates that we are 90%

confident that the true population mean $\mu$ is contained by the interval $(4.1037, 4.9963).$

b) If the population standard deviation, sigma is unknown, then the mean has a student's $t$ distribution and the sample standard deviation is used instead of the population standard deviation.

The maximum error of the estimate is given by the formula

$$E=t_{\alpha/2}\times\dfrac{s}{\sqrt{n}}$$

$$\alpha=0.05, t_{\alpha/2}=2.306002$$

$$E=2.306002\times\dfrac{0.72}{\sqrt{9}}=0.553440$$