Question

the length of time taken on the sats follows a normally distribution with a mean of 2.5 hours and a standard deviation of 0.25 hours. samples of size of 60 are randomly selected from the population.
· x is the normal distribution of the length of time taken on a sat by a student. μ_x = 2.5 and σ_x = 0.25
· n = 60. since x is the normal distribution and n = 60, by clt, we can conclude that (overline{x}) follows a normal distribution.
· (overline{x}sim nleft(2.5,\frac{0.25}{sqrt{60}}
ight))
part a:
find the probability that a sample mean is between two hours and three hours?

Explanation:

Step1: Calculate the z - scores

The formula for the z - score is \(z=\frac{\bar{x}-\mu_{\bar{x}}}{\frac{\sigma_{x}}{\sqrt{n}}}\).
For \(\bar{x} = 2\), \(z_1=\frac{2 - 2.5}{\frac{0.25}{\sqrt{60}}}\)

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For \(\bar{x}=3\), \(z_2=\frac{3 - 2.5}{\frac{0.25}{\sqrt{60}}}\)

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Step2: Find the probability

We know that \(P(2<\bar{X}<3)=P(-15.49 < Z < 15.49)\)
Since the standard normal distribution \(Z\sim N(0,1)\), and for a standard normal random variable \(Z\), \(P(-a<Z < a)\approx1\) when \(a\) is a large number.

Answer:

The probability that a sample mean is between two hours and three hours is approximately \(1\)