A sample size of 500 is sufficiently large enough to conclude that the sampling distribution of the sample proportions is a norm
1 answer:
You might be interested in
Answer:
The estimation for the number of newborns who weighed between 1724 grams and 5172 grams is 595.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
Proportion of newborns who weighed between 1724 grams and 5172 grams.
This is the pvalue of Z when X = 5172 subtracted by the pvalue of Z when X = 1724. So
X = 5172
By the Central Limit Theorem
has a pvalue of 0.9772
X = 1724
has a pvalue of 0.0228
0.9772 - 0.0228 = 0.9544
Estimate the number of newborns who weighed between 1724 grams and 5172 grams.
0.9544 out of 623 babies. SO
0.9544*623 = 595
The estimation for the number of newborns who weighed between 1724 grams and 5172 grams is 595.
Answer: 4x² + 4x - 2 -
<u>Step-by-step explanation:</u>
You can use either synthetic or long division. I am using synthetic division:
2x + 3 = 0 ⇒ x =
| 4 10 4 -6
<u>| ↓ -6 -6 3 </u>
4 4 -2 -3 ← <em>-3 is the remainder</em>
↓ ↓ ↓
4x² +4x -2 ← <em>factored polynomial</em>