Question

this year, the act score of a randomly selected student has an unknown distribution with a mean of 18.9 points and a standard deviation of 4 points. let ( x ) be the act score of a randomly selected student and let ( overline{x} ) be the average act score of a random sample of size 50.

1. describe the probability distribution of ( x ) and state its parameters ( mu ) and ( sigma ):

( x sim ) select an answer ( (mu = 18.9, sigma = 4) )
and find the probability that the act score of a randomly selected student is between 21 and 31 points.
(round the answer to 4 decimal places)

1. use the central limit theorem

the sample size is large ( (n > 30) ) although the distribution of the original population is unknown
to describe the probability distribution of ( overline{x} ) and state its parameters ( mu_{overline{x}} ) and ( sigma_{overline{x}} ): (round the answers to 1 decimal place)
( overline{x} sim n (mu_{overline{x}} = 18.9, sigma_{overline{x}} = 0.6) )
and find the probability that the average act score of a sample of 50 randomly selected students is between 19 and 20 points.
(round the answer to 4 decimal places)

Explanation:

Step1: Calculate z - scores for \(X\)

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\).
For \(x = 21\), \(z_1=\frac{21 - 18.9}{4}=\frac{2.1}{4}=0.525\).
For \(x = 31\), \(z_2=\frac{31 - 18.9}{4}=\frac{12.1}{4}=3.025\).

Step2: Find probabilities using the standard normal distribution

We know that \(P(21<X<31)=P(0.525<Z<3.025)\).
Since \(P(0.525<Z<3.025)=P(Z < 3.025)-P(Z < 0.525)\).
From the standard normal table, \(P(Z < 3.025)\approx0.9987\) and \(P(Z < 0.525)\approx0.6990\).

Step3: Calculate the probability

\(P(0.525<Z<3.025)=0.9987 - 0.6990=0.2997\).

Answer:

\(0.2997\)