Answer to Question #347568 in Statistics and Probability for John

Question #347568

A normally distributed population has a mean of 1,214 and a standard deviation of 122. Find the probability that a single randomly selected element X of the population is between 1,100 and 1,300.


1
Expert's answer
2022-06-03T05:39:22-0400

P(1,100<X<1,300)=P(z1<z<z2),P(1,100<X<1,300)=P(z_1<z<z_2),

where z1 and z2 are z-scores of 1,100 and 1,300.

Let's find them.

z=xμσ;z=\frac{x-\mu}{\sigma}; μ=1,214;\mu=1,214; σ=122.\sigma=122.


z1=1,1001,2141220.93,z_1=\frac{1,100-1,214}{122}\approx-0.93,


z2=1,3001,2141220.70.z_2=\frac{1,300-1,214}{122}\approx0.70.


So we have to rewrite the probability ang use z-table:

P(1,100<X<1,300)=P(0.93<z<0.70)=P(1,100<X<1,300)=P(-0.93<z<0.70)=

=0.75800.1762=0.5818=58.18%.=0.7580-0.1762=0.5818=58.18\%.


Answer: 58.18%.


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