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Given the normally distributed random variable X with mean 18 and variance 6.25, find (a) P(16 < X < 22). (b) the value of k1 such that P(X < k1) = 0.2946 (c) the value of k2 such that P(X > k2) = 0.8186 (d) the two cut-off values that contain the middle 90% of the normal curve area

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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