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In the standard normal z-distribution, is the area between z =-1 and z between z

ID: 3273921 • Letter: I

Question

In the standard normal z-distribution, is the area between z =-1 and z between z 8) 1 the same as the area -0.5 and z = +1.5? why? (Hint: Draw the curve to answer the question). The lifetime of tires is normally distributed with a mean of 50,000 miles and a standard deviation cf 3,000 miles. The warranty is for 46,000 miles. What proportion of the tires will fail after the warranty? 9) 10) If X is a normal random variable with a mean of 20 and a standard deviation of 5, calculate the following probabilities? 1) Probability that X s 12

Explanation / Answer

9) Given that lifetime of tire.

X~ N(mean= 50,000 miles, sd = 3,000 miles)

Also given that the warrenty is for 46,000 miles

Now we have to find P(X > 46,000)

Now convert x = 46,000 into z-score.

z-score is defined as,

z = (x - mean) / sd

z = (46000 - 50000) / 3000 = -1.33

So we have to find P(Z > -1.33)

We can find this probability in EXCEL.

syntax :

=1 - NORMSDIST(z)

where z is z-score.

P(Z > -1.33) = 0.9082

10) Given that,

X ~ N(mean = 20, sd = 5)

Calculate the following probabilities.

1) P(X <=12) :

z-score for x = 12 is,

z = (12 - 20) / 5 = -1.6

That is we have to find P(Z < -1.6).

This probability we can find by using EXCEL.

syntax :

=NORMSDIST(z)

where z is z-score

P(Z < -1.6) = 0.0548

2) P(X >8).

z-score for x = 8 is,

z = (8 - 20) / 5 = -2.4

Now we have to find P(Z > -2.4)

P(Z>-2.4) = 0.9918

3) P(2 <= X <=14)

z-scores for x = 2 and x = 14 are,

z = (2 - 20) / 5 = -3.6

z = (14 - 20)/5 = -1.2

Now we have to find P(-3.6 < Z < -1.2).

P(-3.6 < Z < -1.2) = P(Z < -1.2) - P(Z< -3.6)

= 0.1151 - 0.0002

= 0.1149

11. Given that X ~ N(mean = 50, sd= 8)

Here we have given probability and from this information we have to find x.

P(X > x) = 0.7824

1 - P(X < x) = 0.7824

P(X < x) = 1 - 0.7824

P(X<x) = 0.2176

Now we can find x by using formula,

x = mean + z*sd

where z is z-score for probabiity 0.2176.

z we can find by using EXCEL.

syntax :

=NORMSINV(probability)

where probability = 0.2176

z = -0.78

x = 50 + (-0.78)*8 = 43.76