Answer: 0.07186
Step-by-step explanation:
Let x be a random variable that follows a normal distribution.
Given : Sample size : [tex]n=25[/tex]
Population mean : [tex]\mu=100[/tex]
Standard deviation: [tex]\sigma=10[/tex]
To find : Probability that the sample mean falls in the interval of 1.8 standard deviations from its mean i.e. [tex](\mu-1.8\sigma\ ,\ \mu+1.8 \sigma)[/tex]
z-score: [tex]z=\dfrac{x-\mu}{\sigma}[/tex]
Put [tex]x=\mu-1.8\sigma[/tex]
[tex]z=\dfrac{\mu-1.8\sigma-\mu}{\sigma}=\dfrac{-1.8\sigma}{\sigma}=-1.8[/tex]
Similarly , for [tex]x=\mu+1.8\sigma[/tex]
[tex]z=\dfrac{\mu+1.8\sigma-\mu}{\sigma}=\dfrac{1.8\sigma}{\sigma}=1.8[/tex]
Hence, the probability that the sample mean falls in the interval of 1.8 standard deviations from its mean :-
[tex]P(-1.8<z<1.8)=2(Pz>1.8)=2( 0.0359303)=0.0718606\approx0.07186[/tex]
Hence, the required probability = 0.07186
