Question

suppose that samples of size n 25 are selected at random from a normal population with mean 100 and standard deviation of 10. What is the probability that the sample mean falls in the interval of 1.8 standard deviations from its mean?

Answer 1

Answer: 0.07186

Step-by-step explanation:

Let x be a random variable that follows a normal distribution.

Given : Sample size : $n=25$

Population mean : $\mu=100$

Standard deviation: $\sigma=10$

To find : Probability that the sample mean falls in the interval of 1.8 standard deviations from its mean i.e. $(\mu-1.8\sigma\ ,\ \mu+1.8 \sigma)$

z-score: $z=\dfrac{x-\mu}{\sigma}$

Put $x=\mu-1.8\sigma$

$z=\dfrac{\mu-1.8\sigma-\mu}{\sigma}=\dfrac{-1.8\sigma}{\sigma}=-1.8$

Similarly , for $x=\mu+1.8\sigma$

$z=\dfrac{\mu+1.8\sigma-\mu}{\sigma}=\dfrac{1.8\sigma}{\sigma}=1.8$

Hence, the probability that the sample mean falls in the interval of 1.8 standard deviations from its mean :-

$P(-1.8<z<1.8)=2(Pz>1.8)=2( 0.0359303)=0.0718606\approx0.07186$

Hence, the required probability = 0.07186