Respuesta :
Answer:
(a) The 29th percentile for the number of chocolate chips in a bag is 1198.65.
(b) The number of chocolate chips in a bag that make up the middle 95% of bags are [1146, 1380].
(c) The inter-quartile range of the number of chocolate chips in a bag of chocolate chip cookies is 157.83.
Step-by-step explanation:
Let the random variable X represent the number of chocolate chips in a bag of chocolate chip cookies.
The random variable X is normally distributed with mean, μ = 1263 and a standard deviation, σ = 117.
(a)
Compute the 29th percentile for the number of chocolate chips in a bag as follows:
P (X < x) = 0.29
⇒ P (Z < z) = 0.29
The value of z for the above probability is, z = -0.55.
Compute the value of x as follows:
[tex]z=\frac{x-\mu}{\sogma}\\-0.55=\frac{x-1263}{117}\\x=1263-(117\times 0.55)\\x=1198.65[/tex]
Thus, the 29th percentile for the number of chocolate chips in a bag is 1198.65.
(b)
According to the Empirical rule 95% of the normally distributed data lies within 2 standard deviations of the mean.
P (μ - σ < X < μ + σ) = 0.95
P (1263 - 117 < X < 1263 + 117) = 0.95
P (1146 < X < 1380) = 0.95
Thus, the number of chocolate chips in a bag that make up the middle 95% of bags are [1146, 1380].
(c)
The inter-quartile range of the normal distribution is:
IQR = 1.349 σ
Compute the inter-quartile range of the number of chocolate chips in a bag of chocolate chip cookies as follows:
IQR = 1.349 σ
= 1.349 × 117
= 157.833
Thus, the inter-quartile range of the number of chocolate chips in a bag of chocolate chip cookies is 157.83.
A) The 29th percentile for the number of chocolate chips in a bag is; 1198.65
B )The number of chocolate chips in a bag that make up the middle 95% of bags
C) The interquartile range is; 157.833
What is the interquartile range?
A) Let us calculate the 29th percentile for the number of chocolate chips in a bag as below;
P (Z < z) = 0.29
The z-score for the probability above is; z = -0.55.
To get the 29th percentile, we will make x the subject in the z-score formula to get;
x = zσ + μ
we are given;
σ = 117
μ = 1263
Thus;
x = (-0.55 * 117) + 1263
x = 1198.65
B) From the Empirical rule, 95% of the normally distributed data lies within 2 standard deviations of the mean.
Thus;
P [(μ - σ) < X < (μ + σ)] = 0.95
Plugging in the relevant values;
P[(1263 - 117) < X < (1263 + 117)] = 0.95
P (1146 < X < 1380) = 0.95
Thus, the number of chocolate chips in a bag that make up the middle 95% of bags are between 1146 and 1380 bags
C) The inter-quartile range of the normal distribution is:
IQR = 1.349 σ
Thus;
IQR = 1.349 × 117
IQR = 157.833
Read more about Interquartile Range at; https://brainly.com/question/18723655
