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Sagot :
Answer:
a) 1186
b) Between 1031 and 1493.
c) 160
Step-by-step explanation:
Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula.
In a set 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]
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 p-value, we get the probability that the value of the measure is greater than X.
Normally distributed with mean of 1262 and a standard deviation of 118.
This means that [tex]\mu = 1262, \sigma = 118[/tex]
a) Determine the 26th percentile for the number of chocolate chips in a bag.
This is X when Z has a p-value of 0.26, so X when Z = -0.643.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-0.643 = \frac{X - 1262}{118}[/tex]
[tex]X - 1262 = -0.643*118[/tex]
[tex]X = 1186[/tex]
(b) Determine the number of chocolate chips in a bag that make up the middle 95% of bags.
Between the 50 - (95/2) = 2.5th percentile and the 50 + (95/2) = 97.5th percentile.
2.5th percentile:
X when Z has a p-value of 0.025, so X when Z = -1.96.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-1.96 = \frac{X - 1262}{118}[/tex]
[tex]X - 1262 = -1.96*118[/tex]
[tex]X = 1031[/tex]
97.5th percentile:
X when Z has a p-value of 0.975, so X when Z = 1.96.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.96 = \frac{X - 1262}{118}[/tex]
[tex]X - 1262 = 1.96*118[/tex]
[tex]X = 1493[/tex]
Between 1031 and 1493.
(c) What is the interquartile range of the number of chocolate chips in a bag of chocolate chip cookies?
Difference between the 75th percentile and the 25th percentile.
25th percentile:
X when Z has a p-value of 0.25, so X when Z = -0.675.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-0.675 = \frac{X - 1262}{118}[/tex]
[tex]X - 1262 = -0.675*118[/tex]
[tex]X = 1182[/tex]
75th percentile:
X when Z has a p-value of 0.75, so X when Z = 0.675.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]0.675 = \frac{X - 1262}{118}[/tex]
[tex]X - 1262 = 0.675*118[/tex]
[tex]X = 1342[/tex]
IQR:
1342 - 1182 = 160
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