Question

Given a normal distribution with μ=40 and σ=66​, find​ (a) the normal curve area to the right of x=24; (b) the normal curve area to the left of x=29; (c) the normal curve area between x=47 and x=54; ​(d) the value of x that has 70​% of the normal curve area to the​ left; and​ (e) the two values of x that contain the middle 65​% of the normal curve area.

Answer 1

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Solution :-

Given :-

μ=40 and σ=66

a)We have to find P(X>24)

Using z-score,

P(Z>−0.2424) = 1-P(Z<−0.2424)

P(Z>−0.2424) = 1−0.4042=0.5958 Using standard normal table.

Hence, P(X>24) = 0.5958

b)

We have to find P(X<29)

Using z-score,

P(Z≤−0.1667)=0.4338 Using standard normal table.

Hence, P(X<29) = 0.4338

c)

We have to find P(47<X<54) = ...?

Using z-score,

P(0.1061≤Z≤0.2121) = P(Z≤0.2121)−Pr(Z≤0.1061)

=0.584−0.5422=0.0418 ...Using standard normal table

Hence, P(47<X<54) = 0.0418

d)

We have to find value of x such that P(Z<z) = 0.70

Therefore, P(Z<0.5244) = 0.70 ...Using standard normal table,

So, z = 0.5244

Using formula,

μ=40 and σ=66

x=μ+z* σ

x = 40+0.5244*66

x=74.61

Done