Respuesta :
Answer:
(a) P(1000 < X < 1500) = 0.0256
(b) P(X < 1025) = 0.0392
(c) P(X > 1200) = 0.6554
(d) Percentile rank of a bag that contains 1425 chocolate chips = 90.98%
Step-by-step explanation:
We are given that the number of chocolate chips in an 18-ounce bag of chocolate chip cookies is approximately normally distributed with a mean of 1252 chips and standard deviation 129 chips.
Firstly, Let X = number of chocolate chips in a bag
The z score probability distribution for is given by;
Z = [tex]\frac{ X - \mu}{\sigma}[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = population mean = 1252 chips
[tex]\sigma[/tex] = standard deviation = 129 chips
(a) Probability that a randomly selected bag contains between 1000 and 1500 chocolate chips, inclusive is given by = P(1000 [tex]\leq[/tex] X [tex]\leq[/tex] 1500) = P(X [tex]\leq[/tex] 1500) - P(X < 1000)
P(X [tex]\leq[/tex] 1500) = P( [tex]\frac{ X - \mu}{\sigma}[/tex] [tex]\leq[/tex] [tex]\frac{1500-1252}{129}[/tex] ) = P(Z [tex]\leq[/tex] 1.92) = 0.9726
P(X < 1000) = P( [tex]\frac{ X - \mu}{\sigma}[/tex] < [tex]\frac{1000-1252}{129}[/tex] ) = P(Z < -1.95) = 1 - P(Z [tex]\leq[/tex] 1.95)
= 1 - 0.9744 = 0.0256
Therefore, P(1000 [tex]\leq[/tex] X [tex]\leq[/tex] 1500) = 0.9726 - 0.0256 = 0.947
(b) Probability that a randomly selected bag contains fewer than 1025 chocolate chips is given by = P(X < 1025)
P(X < 1025) = P( [tex]\frac{ X - \mu}{\sigma}[/tex] < [tex]\frac{1025-1252}{129}[/tex] ) = P(Z < -1.76) = 1 - P(Z [tex]\leq[/tex] 1.76)
= 1 - 0.9608 = 0.0392
(c) Proportion of bags contains more than 1200 chocolate chips is given by = P(X > 1200)
P(X > 1025) = P( [tex]\frac{ X - \mu}{\sigma}[/tex] > [tex]\frac{1200-1252}{129}[/tex] ) = P(Z > -0.40) = P(Z < 0.40) = 0.6554
(d) Percentile rank of a bag that contains 1425 chocolate chips is given by;
Firstly we will calculate the z score of 1425 chocolate chips, i.e.;
Z = [tex]\frac{1425-1252}{129}[/tex] = 1.34
Now, we will check the area probability in z table which corresponds to this critical value of x;
The value which we get is 0.9098.
Therefore, 90.98% is the rank of bag that contains 1425 chocolate chips.
Using the normal distribution, it is found that:
a) There is a 0.947 = 94.7% probability that a randomly selected bag contains between 1000 and 1500 chocolate chips, inclusive.
b) There is a 0.0392 = 3.92% probability that a randomly selected bag contains fewer than 1025 chocolate chips.
c) 0.3446 = 34.46% of bags contains more than 1200 chocolate chips.
d) The bag is in the 91st percentile.
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.
- 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 1252 chips, thus [tex]\mu = 1252[/tex].
- Standard deviation of 129 chips, thus [tex]\sigma = 129[/tex].
Item a:
The probability is the p-value of Z when X = 1500 subtracted by the p-value of Z when X = 1000, thus:
X = 1500:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{1500 - 1252}{129}[/tex]
[tex]Z = 1.92[/tex]
[tex]Z = 1.92[/tex] has a p-value of 0.9726.
X = 1000:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{1000 - 1252}{129}[/tex]
[tex]Z = -1.95[/tex]
[tex]Z = -1.95[/tex] has a p-value of 0.0256.
Then, 0.9726 - 0.0256 = 0.947
There is a 0.947 = 94.7% probability that a randomly selected bag contains between 1000 and 1500 chocolate chips, inclusive.
Item b:
This probability is the p-value of Z when X = 1025, thus:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{1025 - 1252}{129}[/tex]
[tex]Z = -1.76[/tex]
[tex]Z = -1.76[/tex] has a p-value of 0.0392.
There is a 0.0392 = 3.92% probability that a randomly selected bag contains fewer than 1025 chocolate chips.
Item c:
This proportion is 1 subtracted by the p-value of Z when X = 1200, thus:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{1200 - 1252}{129}[/tex]
[tex]Z = -0.4[/tex]
[tex]Z = -0.4[/tex] has a p-value of 0.3446.
0.3446 = 34.46% of bags contains more than 1200 chocolate chips.
Item d:
This percentile is the p-value of Z when X = 1425, thus:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{1425 - 1252}{129}[/tex]
[tex]Z = 1.34[/tex]
[tex]Z = 1.34[/tex] has a p-value of 0.91.
The bag is in the 91st percentile.
A similar problem is given at https://brainly.com/question/13680644
