Answer: 0.6827
Step-by-step explanation:
Given : Mean IQ score : ![\mu=105](https://tex.z-dn.net/?f=%5Cmu%3D105%20)
Standard deviation : ![\sigma=15](https://tex.z-dn.net/?f=%5Csigma%3D15)
We assume that adults have IQ scores that are normally distributed .
Let x be the random variable that represents the IQ score of adults .
z-score : ![z=\dfrac{x-\mu}{\sigma}](https://tex.z-dn.net/?f=z%3D%5Cdfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D)
For x= 90
![z=\dfrac{90-105}{15}\approx-1](https://tex.z-dn.net/?f=z%3D%5Cdfrac%7B90-105%7D%7B15%7D%5Capprox-1)
For x= 120
![z=\dfrac{120-105}{15}\approx1](https://tex.z-dn.net/?f=z%3D%5Cdfrac%7B120-105%7D%7B15%7D%5Capprox1)
By using the standard normal distribution table , we have
The p-value : ![P(90](https://tex.z-dn.net/?f=P%2890%3Cx%3C120%29%3DP%28-1%3Cz%3C1%29)
![P(z](https://tex.z-dn.net/?f=P%28z%3C1%29-P%28z%3C-1%29%3D%200.8413447-0.1586553%3D0.6826894%5Capprox0.6827)
Hence, the probability that a randomly selected adult has an IQ between 90 and 120 =0.6827
Answer:
just doing this for my points!!! lol
Step-by-step explanation:
Answer:
y=35x+50
35*8=280
280+50=330
yes there will be 10 dollars left
I hope this is good enough:
Answer:
3.84% probability that it has a low birth weight
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:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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:
![\mu = 3466, \sigma = 546](https://tex.z-dn.net/?f=%5Cmu%20%3D%203466%2C%20%5Csigma%20%3D%20546)
If we randomly select a baby, what is the probability that it has a low birth weight?
This is the pvalue of Z when X = 2500. So
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{2500 - 3466}{546}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B2500%20-%203466%7D%7B546%7D)
![Z = -1.77](https://tex.z-dn.net/?f=Z%20%3D%20-1.77)
has a pvalue of 0.0384
3.84% probability that it has a low birth weight
Interest=1250-880=370
I=PRT
P=principal
r=rate in deicmal
t=time in years
1 month so t=1/12
P=880
I=370
370=880*r*1/12
370=220/3r
times both sides by 3
1110=220r
divdie both sides by 220
5.04545=r
505% interest