Question

the shelf life of a particular dairy product is normally distributed with a mean of 12 days and a standard deviation of 3 days. About what percent of the products last between 12 and 15 days?

Answer 1

Answer:

a) P(9<X<15)=P(\frac{9-12}{3}<Z<\frac{15-12}{3})=P(-1<Z<1)=0.6826=68.26\%.P(9<X<15)=P(39−12<Z<315−12)=P(−1<Z<1)=0.6826=68.26%.

b) P(12<X<15)=P(\frac{12-12}{3}<Z<\frac{15-12}{3})=P(0<Z<1)=0.3413=34.13\%.P(12<X<15)=P(312−12<Z<315−12)=P(0<Z<1)=0.3413=34.13%.

c) P(<6)=P(Z<\frac{6-12}{3})=P(Z<-2)=0.0228=2.28\%.P(<6)=P(Z<36−12)=P(Z<−2)=0.0228=2.28%.

Answer 2

Answer:

$55 | < \sqrt[x {12 = \leqslant \leqslant < {)x { {( { \frac{ \frac{ \frac{ \sqrt[45 = ]{?} }{?} }{?} }{?} } \cot( \beta \alpha log_{?}(?) ) ^{2} }^{2} }^{2} }^{?} }^{?} ]{?} |$

Step-by-step explanation:

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