91Ó°ÊÓ

According to the business that packages the salt, each bag contains an average of $40\mathrm{lb}$ of salt, with a standard deviation of $1.5\mathrm{lb}$

Formula used: Standard deviation${\mathrm{σ}}_{\stackrel{¯}{X}}=\frac{\mathrm{σ}}{\sqrt{n}}$

Under the assumption that weights are normally distributed with $\mathrm{μ}=40\mathrm{lb}\&\mathrm{σ}=1.5\mathrm{lb}$, we must find the chance that a randomly selected bag will be $39\mathrm{lb}$or less.

We have to find $P(X≤39)$Where $X~N\left(40,1.5^2\right)\left[\begin{array}{l}\text{Where}X\text{is the random}\\ \text{variable denoting}\\ \text{the weight of a bag}\end{array}\right.$

$$\begin{gathered} ∴P(X≤39) \\ =P\left(\frac{X-\mathrm{μ}}{\mathrm{σ}}≤\frac{39-\mathrm{μ}}{\mathrm{σ}}\right) \\ =P\left(z≤\frac{39-40}{1.5}\right)\left[z=\frac{X-\mathrm{μ}}{\mathrm{σ}},\mathrm{μ}=40\mathrm{σ}=1.5,∴z~N(0,1)\right. \\ =P\left(z≤-\frac{1}{1.5}\right) \\ =P(z≤-0.66667) \end{gathered}$$

$$\begin{gathered} =\mathrm{Φ}(-0.66667)\left[\begin{array}{r}\text{where}\mathrm{Φ}\left(u\right)=P(z≤u)\\ =c·d·f\text{of}z\end{array}\right. \\ =1-\mathrm{Φ}(0.66667) \\ [âˆ\mu \mathrm{Φ}(-u)=1-\mathrm{Φ}(u\left)\right] \\ =1-0.7475075 \\ =0.2524925 \end{gathered}$$

We've picked a random sample of ten bags (sample size $n=10$).

For samples of size $10$, the sample mean $\stackrel{¯}{x}$now follows a normal distribution with a mean $\mathrm{μ}=40\mathrm{lb}$& S.D ${\mathrm{σ}}_{\stackrel{¯}{X}}=\frac{\mathrm{σ}}{\sqrt{n}}$

$$\begin{gathered} =\frac{1.5}{\sqrt{10}}\mathrm{lb} \\ =.47434\mathrm{lb} \end{gathered}$$

We have to find, $P(\stackrel{¯}{x}≤39)$

$$\begin{gathered} P(\stackrel{¯}{x}≤39) \\ =P\left(\frac{\stackrel{¯}{x}-\mathrm{μ}}{{\mathrm{σ}}_{\stackrel{¯}{X}}}≤\frac{39-\mathrm{μ}}{{\mathrm{σ}}_{\stackrel{¯}{X}}}\right) \\ =P\left(z≤\frac{39-40}{.47434}\right)\text{Where}z=\frac{\stackrel{¯}{x}-\mathrm{μ}}{{\mathrm{σ}}_{\stackrel{¯}{X}}}~N(0,1) \\ =P(z≤-2.10818) \\ =\mathrm{Φ}(-2.10818)\left[\begin{array}{r}\mathrm{Φ}\left(u\right)=P(z≤u)\\ =c·d·f\text{of}z\end{array}\right. \\ =1-\mathrm{Φ}(2.10818)\left[\text{From the table of}\mathrm{Φ}\right(u)\text{of standard normal} \\ =1-0.982423\text{variable}\mathrm{Φ}(2.10818)=0.9824923 \\ =0.0175077 \end{gathered}$$

No, this does not suggest that the firm's assertion is false. The manufacturer states that the bag weights are regularly distributed, with a mean of $40\mathrm{lbs}\&$a standard deviation of $\mathrm{σ}=1.5\mathrm{lbs}$This does not imply that all of the bags are $40\mathrm{lb}$in weight. If all of the bags are $40\mathrm{lb}$, the weight of the bags will follow a degenerate distribution degenerated at $40\mathrm{lb}$rather than a normal distribution. What company states that a large number of bags have been clustered around the weight (average weight) of $40\mathrm{lb}$? The claim "S.D of the weights is $1.5\mathrm{lb}$" indicates that the weights differ from the average weight in some way. $40\mathrm{b}$

As a result, some variation in population weight is natural. So, when we bought a bag, that is, when we randomly selected or sampled a bag, it is possible that the bag weighs $39\mathrm{lb}$ due to chance, but based on the calculated probability of having a bag weighing $39\mathrm{lb}$ or less $(.25)$, we can say that the chance of having that bag is low, which is exactly what the company claimed.

No, deciding that the company's claim is false based on the fact that the mean weight for a sample of $10$ is $39\mathrm{lb}$ is not correct. Because the sample size is so tiny, substantial sampling errors due to chance are more likely.

If we need to infer something about the population mean based on this sample, we usually test some hypothesis about the population mean's value of interest or find a confidence interval for the population mean.