let's firstly, convert the mixed fraction to improper fraction, and then subtract.
![\bf \stackrel{mixed}{2\frac{3}{4}}\implies \cfrac{2\cdot 4+3}{4}\implies \stackrel{improper}{\cfrac{11}{4}} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \cfrac{11}{4}-\cfrac{2}{3}\implies \stackrel{\textit{our LCD will be 12}}{\cfrac{(3)11-(4)2}{12}}\implies \cfrac{33-8}{12}\implies \cfrac{25}{12}\implies 2\frac{1}{12}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7Bmixed%7D%7B2%5Cfrac%7B3%7D%7B4%7D%7D%5Cimplies%20%5Ccfrac%7B2%5Ccdot%204%2B3%7D%7B4%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B11%7D%7B4%7D%7D%0A%5C%5C%5C%5C%5B-0.35em%5D%0A%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%0A%5Ccfrac%7B11%7D%7B4%7D-%5Ccfrac%7B2%7D%7B3%7D%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bour%20LCD%20will%20be%2012%7D%7D%7B%5Ccfrac%7B%283%2911-%284%292%7D%7B12%7D%7D%5Cimplies%20%5Ccfrac%7B33-8%7D%7B12%7D%5Cimplies%20%5Ccfrac%7B25%7D%7B12%7D%5Cimplies%202%5Cfrac%7B1%7D%7B12%7D)
Answer:
a) 281 days.
b) 255 days
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:

(a) What is the minimum pregnancy length that can be in the top 8% of pregnancy lengths?
100 - 8 = 92th percentile.
X when Z has a pvalue of 0.92. So X when Z = 1.405.




(b) What is the maximum pregnancy length that can be in the bottom 3% of pregnancy lengths?
3rd percentile.
X when Z has a pvalue of 0.03. So X when Z = -1.88




=
5
−
4 Does the answer that’s what I got on my quiz I just took the test