Question

A random variable x is normally distributed:  x~N(μ=74, σ=4.3).

What percent of the population values will be greater than 77.9?

Enter in percent form (without %), correct to two digits after the decimal point:   

We want to change μ, without changing σ, such that in this new distribution, 30% of the values would be higher than 77.9. Determine the new value of μ.

Give the answer correct to two digits after the decimal point:

Answer 1

Solution :

Given that ,

a) P(x > 77.9) = 1 - p( x< 77.9)

=1- p P[(x - ) / < (77.9 - 74) / 4.3 ]

=1- P(z < 0.91)

= 1 - 0.8186

= 0.1814

percentage = 18.14%

b) Using standard normal table,

P(Z > z) = 30%

= 1 - P(Z < z) = 0.30  

= P(Z < z) = 1 - 0.30

= P(Z < 0.524 ) = 0.70  

z = 0.524

Using z-score formula,

x = z * +

77.9 = 0.52 * 4.3 +

= 77.9 - 0.524 * 4.3

= 75.65