Question

The scores on a certain test are normally distributed with a mean score of 43 and a standard deviation of 3. What is the probability that a sample of 90 students will have a mean score of at least 43.3162?

0.8413

0.3174

0.1587

0.3413

Answer 1

Answer is C, hope this helps. If you have more questions, post them in a forum, and I will be happy to answer. -Haяяison

Answer 2

Answer: 0.1587

Explanation:

Given: Mean :
`\mu=43`

Standard deviation :
`\sigma=3`

Sample size : n = 90

We know that , the value of z is given by :-

`z=(x-\mu)/((\sigma)/(√(n)))`

For x=4.2

`z=(43.3162-43)/((3)/(√(90)))=0.999912196145\approx1`

The p value =
`P(Z\geq1)=1-P(z<1)=1- 0.8413447= 0.1586553\approx0.1587`

Hence, the probability that a sample of 90 students will have a mean score of at least 43.3162= 0.1587