Given:
Sample Mean <span>= 30<span>
Sample size </span><span><span><span>= 1000</span></span><span>
</span></span></span>Population Standard deviation or <span><span><span>σ<span>=2</span></span><span>
</span></span>Confidence interval </span><span>= 95%</span>
to compute for the confidence interval
Population Mean or <span>μ<span><span>= sample mean ± (</span>z×<span>SE</span>)</span></span>
<span><span>where:</span></span>
<span><span>SE</span>→</span> Standard Error
<span><span>SE</span>=<span>σ<span>√n</span>= 30</span></span>√1000=0.9486
Critical Value of z for 95% confidence interval <span>=1.96</span>
<span>μ<span>=30±<span>(1.96×0.9486)</span></span><span>
</span></span><span>μ<span>=30±1.8594</span></span>
Upper Limit
<span>μ <span>= 30 + 1.8594 = 31.8594</span></span>
Lower Limit
<span>μ <span>= 30 − 1.8594 = <span>28.1406</span></span></span>
<span><span><span>
</span></span></span>
<span><span><span>answer: 28.1406<u<31.8594</span></span></span>
We need the rest of the question :)
Answer:ygggydtdgdgdfddgfs
Step-by-step explanation:
Answer:
-15
Step-by-step explanation:
-3+8-5. 5x3=15 and because it is -p it would be -15.
Answer:
0.13591.
Step-by-step explanation:
We re asked to find the probability of randomly selecting a score between 1 and 2 standard deviations below the mean.
We know that z-score tells us that a data point is how many standard deviation above or below mean.
To solve our given problem, we need to find area between z-score of -2 and -1 that is 
.
We will use formula 
to solve our given problem.
Using normal distribution table, we will get:
Therefore, the probability of randomly selecting a score between 1 and 2 standard deviations below the mean would be 0.13591.
