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The mass of certain types of cabbage is normally distributed with a mean of 1000 g and a standard deviation of 0.15 kg. In a batch of 800 cabbages, estimate how many a mass between 750 g and 1290 g.

Final Answer:

x =  740

Step-by-step explanation:

$$\begin{gathered} \mu =1000\text{g} \\ \sigma =0.15\text{kg}\to \text{g}=0.15\cdot 1000\text{g}=150\text{g} \\ n=800 \\ {M}_1=750\text{g} \\ {M}_2=1290\text{g} \\ p=M_1<N(\mu ,\sigma )<M_2 \\ {k}_1=(\mu -M_1)/\sigma =(1000-750)/150=\frac{5}{3}=1\frac{2}{3}\doteq 1.6667 \\ {k}_2=(M_2-\mu )/\sigma =(1290-1000)/150=\frac{29}{15}=1\frac{14}{15}\doteq 1.9333 \\ K=k_1+k_2=1.6667+1.9333=\frac{18}{5}=3\frac{3}{5}=3.6 \\ p=0.9256 \\ {x}_1=n\cdot p=800\cdot 0.9256=\frac{18512}{25}=740\frac{12}{25}=740.48 \\ x=\lfloor x_1\rfloor =\lfloor 740.48\rfloor =740 \end{gathered}$$

Use the normal distribution table.

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