Respuesta :
Answer:
0.13591 = 13.59%.
Step-by-step explanation:
We have been given that weekly wages at a certain factory are normally distributed with a mean of $400 and a standard deviation of $50.
First of all let us find the z-score for our given sample score using z-score formula.
[tex]z=\frac{x-\mu}{\sigma}[/tex], where,
[tex]z=\text{z-score}[/tex],
[tex]x=\text{Sample-score}[/tex],
[tex]\mu=\text{Mean}[/tex],
[tex]\sigma=\text{Standard deviation}[/tex].
[tex]z=\frac{450-400}{50}[/tex]
[tex]z=\frac{50}{50}[/tex]
[tex]z=1[/tex]
Let us find z-score for sample score 500.
[tex]z=\frac{500-400}{50}[/tex]
[tex]z=\frac{100}{50}[/tex]
[tex]z=2[/tex]
Let us find the probability of both z-score using normal distribution table.
[tex]P(z<1)=0.84134 [/tex]
[tex]P(z<2)=0.97725[/tex]
Since we know that probability between two z-scores can be found by subtracting the smaller area from the larger area as:
[tex]P(1<z<2)=P(z<2)-P(z<1)[/tex]
Upon substituting our values we will get,
[tex]P(1<z<2)=0.97725-0.84134[/tex]
[tex]P(1<z<2)=0.13591[/tex]
Therefore, the probability that a worker selected at random makes between $450 and $500 is 0.13591 or 13.59%.
