The mean of a normally distributed dataset is 12, and the standard deviation is 2. % of the data points lies between 8 and 16.

Question

asked Jul 3, 2018 137k views

2 Answers

BlackXero · Answer 1 · 2018-07-08T11:50:37+0000

Answer:

95.44% of the data lies between 8 and 16.

Explanation:

Since, z score or standard score formula is,

$z=(x-\mu)/(\sigma)$

Where,

$\mu$ = mean of the data,

$\sigma$ = standard deviation,

Let X represents a data point,

So, we have to find out,

P( 8 < X < 16),

Since,

$P(8 < X < 16)=P((8-\mu)/(\sigma)< Z< (16-\mu)/(\sigma))$

=P((8-12)/(2)<Z<(16-12)/(2))

=P(-2<Z<2)

=P(Z<2) - P(Z<-2)

=0.9772-0.0228 ( By the z-score table )

=0.9544

$=95.44\%$

Hence, 95.44% of the data lies between 8 and 16.