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Answer :
Answer: 0.0047
Step-by-step explanation:
Given : A manufacturer knows that their items have a normally distributed length, with a mean of 7.1 inches, and standard deviation of 1.7 inches.
i.e. [tex]\mu=7.1\text{ inches}[/tex]
[tex]\sigma=17\text{ inches}[/tex]
Sample size : n= 24
Let [tex]\overline{X}[/tex] be the sample mean.
Formula : [tex]z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
Then, the probability that their mean length is less than 6.2 inches will be :-
[tex]P(\overline{x}<6.2)=P(\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}<\dfrac{6.2-7.1}{\dfrac{1.7}{\sqrt{24}}})\\\\\approx P(z<-2.6)\\\\=1-P(z<2.6)\ \ [\because\ P(Z<-z)=1-P(Z hence,. the required probability = 0.0047
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