If the amount of change a statistics student carries has a mean of $0.88 and a standard deviation of $0.30, suppose that we randomly pick 36 online statistics students. Find the probability the average of 36 students is between $.80 and $1. The Probability that an individual student has between $.80 and $1 is .9370.

Respuesta :

We have the following:

We must calculate the value of z

a.

[tex]Z=\frac{x-m}{s}[/tex]

x, is the value to evalute (0.8 - 1)

m, is the mean (0.88)

s, is the standard deviation (0.3)

n, is the sample size (36)

[tex]\begin{gathered} \frac{0.8-0.88}{0.3}Now,[tex]0.6554-0.3974=0.258[/tex]

Therefore, the probability is 0.258 or 25.8%

b.

[tex]Z=\frac{x-m}{\frac{s}{\sqrt[]{n}}}[/tex]

x, is the value to evalute (0.8 - 1)

m, is the mean (0.88)

s, is the standard deviation (0.3)

n, is the sample size (36)

replacing:

[tex]\begin{gathered} \frac{0.8-0.88}{\frac{0.3}{\sqrt[]{36}}}now,[tex]0.9918-0.0548=0.937[/tex]

Therefore, the probability is 0.937 or 93.7%

Ver imagen JalexiK148894
Ver imagen JalexiK148894