Westonci.ca is the ultimate Q&A platform, offering detailed and reliable answers from a knowledgeable community. Join our Q&A platform and connect with professionals ready to provide precise answers to your questions in various areas. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.
Sagot :
for the given normally distributed random variable,
(a) P(16 < X < 22) = 0.7333
(b) P(X < k1) = 0.2946; such that the value of k1 = 16.625
(c) P(X > k2) = 0.8186; such that the value of k2 = 15.725
(d) The two cut-off values that contain the middle 90% of the normal curve area are -1.64 and +1.64.
What is the formula for finding the z-score?
The formula for finding the z-score is
z = (X - μ)/σ
Here the mean (μ) and the standard deviation (σ) are applied.
Calculation:
It is given that, a normally distributed random variable X has a mean μ = 18 and the variance σ² = 6.25.
So, its standard deviation is σ = √6.25 = 2.5
(a) P(16 < X < 22)
The z-score for X = 16 is
z = (X - μ)/σ ⇒ z = (16 - 18)/2.5 = -0.8
The z-score for X = 22 is
z = (X - μ)/σ ⇒ z = (22 - 18)/2.5 = 1.6
So, the probability becomes
P(16 < X < 22) = P(-0.8 < z < 1.6)
            = P(z < 1.6) - P(z < -0.8)
From the distribution table, we have P(z < 1.6) = 0.9452 and P(z < -0.8) = 0.2119.
So, we get
P(16 < X < 22) = 0.9452 - 0.2119 = 0.7333
(b) The value of k1 such that P(X < k1) = 0.2946:
we have P(X < k1) = 0.2946
⇒ P(X < k1) = P(z < (k1 - μ)/σ) = 0.2946
From the table, we know that,
P(z < -0.55) = 0.2946
So,
(k1 - μ)/σ = -0.55
⇒ (k1 - 18)/2.5 = -0.55
⇒ k1 - 18 = -0.55 × 2.5 = -1.375
∴ k1 = -1.375 + 18 = 16.625
(c) The value of k2 such that P(X > k2) = 0.8186:
we have P(X > k2) = 0.8186
⇒ P(z > (k2 - μ)/σ) = 0.8186
⇒ 1 - P(z < (k2 - μ)/σ) = 0.8186
⇒ P(z < (k2 - μ)/σ) = 1 - 0.8186 = 0.1814
From the table, we know that P(z < -0.91) = 0.1814
So,
(k2 - μ)/σ = -0.91
⇒ (k2 - 18)/2.5 = -0.91
⇒ k2 - 18 = -0.91 × 2.5 = -2.275
∴ k2 = -2.275 + 18 = 15.725
(d) The two cut-off values that contain the middle 90% of the normal curve area are -1.64 and +1.64 since the z-score for a 90% confidence interval is 1.64.
Learn more about the normal distribution here:
https://brainly.com/question/26822684
#SPJ4
Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.
