If someone could answer this problem and explain how to do it that would be a HUGE help thanks-
2 answers:
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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 <em>X</em> represent the number of chocolate chips in a bag of chocolate chip cookies.
The random variable <em>X</em> is normally distributed with mean, <em>μ </em>= 1263 and a standard deviation, <em>σ </em>= 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 <em>z</em> for the above probability is, <em>z</em> = -0.55.
Compute the value of <em>x</em> as follows:
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 <em>σ</em>
Compute the inter-quartile range of the number of chocolate chips in a bag of chocolate chip cookies as follows:
IQR = 1.349 <em>σ</em>
= 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.