1- 2- If X, . . . , X22 is a random sample from a normal...

Question

Question

1-

If X, . . . , X22 is a random sample from a normal distribution with parameters μ = 18.4 and ơa_ 13.3, find P(164.11-Σ(x-18.4)2-535.86)

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math Statistics-And-Probability

Answer 1

Answer #1

Question 1 The given question represents the chi-square distibution in definition. Here,

the value of $\sum_{i=1}^{22}(X_i - 18.4)^2$ = Z_i² * σ²

so as we know that

X²_i = $\sum_{i=1}^{k}Z_i^2$

so here k = 22 degree of freedom and we can rewrite the given probability as

P(164.11/13.3 < $\sum_{i=1}^{22}Z_i^2$ < 535.86/13.3) = P(12.339 < $\sum_{i=1}^{k}Z_i^2$ < 40.365)

= 0.95 - 0.01 = 0.94

as these probability can be calculated by CHIDIST formula or chi - square table .

(2) Here Pr(right handed) = 0.80

Pr(Left handed) = 0.16

Pr(Ambidextrous) = 0.04

Pr(121 RH, 22 LH , 7 ambi) = ¹⁵⁰C₁₂₁ (0.80)¹²¹* ²⁹C₂₂ (0.16)²² * ⁷C₇ (0.04)⁷ = 0.01187

Answer 2

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