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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
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