The scores on a certain test are normally distributed with a mean score of 69 and a standard deviation of 4. what is the probability that a sample of 90 students will have a mean score of at least 69.4216? your answer should be a decimal rounded to the fourth decimal place.

Solution: We are given:

μ=69,σ=4,n=90

We have to find P(Mean≥69.4216)

We need to first find the z score

z= (xbar-μ)/(σ/sqrt(n))

=(69.4216-69)/(4/sqrt(90))

=1.00

Now we have to find P(z≥1.00)

Using the standard normal table, we have:

P(z≥1.00)= 0.1587

Therefore the probability that a sample of 90 students will have a mean score of at least 69.4216 is 0.1587.

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