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

Step-by-step explanation:

Given: Mean :$\mu=43$

Standard deviation : $\sigma=3$

Sample size : n = 90

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

$z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}$

For x=4.2

$z=\dfrac{43.3162-43}{\dfrac{3}{\sqrt{90}}}=0.999912196145\approx1$

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