Respuesta :
Answer:
The probability that a randomly selected chocolate bar will have a. Between 200 and 220 calories will be 0.3023
Step-by-step explanation:
Given:
True mean=225
S.D=10
X1=200 and X2=220
To Find:
P(X1<x<X2)
Solution:
BY using Z-table for probability and Z-score we can proceed.
So
For 200 calories Z will be ,
Z=(sample mean -true mean)/S.D
=200-225/10
=-25/10
=-2.5
For 220 calories Z will be ,
Z=220-225/10
=-5/10
=-0.5
So required probability will be
[tex]P(-2.5<z<-0.5)[/tex]
[tex]=P(-2.5<z<0)-P(-0.5<z<0)[/tex]
[tex]=P(0>z>2.5)-P(0>z>0.5)[/tex]
[tex]=0.4938-0.1915[/tex]
=0.3023
Using the normal distribution, it is found that there is a 0.3023 = 30.23% probability that a randomly selected chocolate bar will have between 200 and 220 calories.
In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- It measures how many standard deviations the measure is from the mean. Â
- After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
In this problem:
- Mean of 225 calories, thus [tex]\mu = 225[/tex].
- Standard deviation of 10 calories, thus [tex]\sigma = 10[/tex].
The probability that a chocolate bar will have between 200 and 220 calories is the p-value of Z when X = 220 subtracted by the p-value of Z when X = 200, thus:
X = 220:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{220 - 225}{10}[/tex]
[tex]Z = -0.5[/tex]
[tex]Z = -0.5[/tex] has a p-value of 0.3085.
X = 200:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{200 - 225}{10}[/tex]
[tex]Z = -2.5[/tex]
[tex]Z = -2.5[/tex] has a p-value of 0.0062.
0.3085 - 0.0062 = 0.3023.
0.3023 = 30.23% probability that a randomly selected chocolate bar will have between 200 and 220 calories.
A similar problem is given at https://brainly.com/question/24663213
