Question

Scores on the math portion of the SAT (SAT-M) in a recent year have followed a normal distribution with mean μ = 507 and standard deviation σ = 111.

What is the probability that the mean SAT-M score of a random sample of 4 students who took the test that year is more than 600?

If normality conditions are met, round your Z-score calculation to 2 decimal places.

Normal Distribution Calculator (Standard Normal)

Group of answer choices

1. P(Z>1.68) = 0.05

2. P(Z>0.84)=0.20P(Z>0.84)=0.20P(Z>0.84)=0.20P(Z>0.84)=0.20

3. It is impossible to tell because normality conditions are not met

Answer 1

It is given that scores on the math portion of the SAT (SAT-M) in a recent year have followed a normal distribution with mean μ = 507 and standard deviation σ = 111.

Further, let
1. , X2, X3, X4
denote the SAT-M score of a random sample of 4 students. Then
X;" N (507,1112); i = 1,2,3,4

Now we know that if $\small X_1, \dots, X_n$ is a random sample from $\small \mathcal{N}(\mu, \sigma^2)$ distribution, then the sample mean $\small \bar{X}$ also follows Normal distribution with the same mean but a scaled variance, that is $\small \bar{X} \sim \mathcal{N}(\mu, \frac{\sigma^2}{n})$ .

Thus, if
1 X = (X1 + X2 + X3 + X4)
is the mean score of the random sample of 4 students, then,
111? XN (507,
. Hence, the normality assumption holds.

Now, we want to calculate . Towards that, we calculate the z-score as,

P(X > 600

X 600 - 507 93 * 2 111 1.6757 1112

Rounding off to 2-decimal places, we get z = 1.68. Thus, the required probability is

Hence, option (A) is correct.

P(X > 600) = P(Z >=) = P(Z > 1.68)

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111? XN (507,

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