Question

the mean height of a group of 500 nonsmoking college students is 74 inches and the standard deviation is 5 inches. what is the probability that in a random sample of 25 students from this group, the average height will be between 73 and 75 inches?

Answer 1

Yes I believe

Answer 2

Wait. Nevermind.

Answer 3

??

Answer 4

www.mathportal.org/calculators/statistics-calculator/normal-distribution-calculator.php?val1=74&val2=5&rb1=s&val3=73&val4=75&ch1=expl

Answer 5

Hey there.

Answer 6

In this type of exercise, you have to know whether you use normal or binomial distributions. Such thing, is determined by seeing the value of np and nq, where n is the number of trials, p is the probability of success of a trial and q is the probability of failure of a trial.
Before we go to calculations, take note that if either or both np and nq are less than 5, then a binomial expansion is to be used. Otherwise, a normal approximation is to be used.
NB: \[np=mean=\mu \] and\[q=1-p\]
according to the information given, np=74, with n as 500.\[p=\frac{ \mu }{ n }=\frac{ 74 }{ 500 }=0.148\]\[nq=n(1-p)=500(1-0.148)=426\]

Conclusively, a normal approximation is to be used.

Answer 7

THe variables needed in a normal distribution are mean, standard deviation and X-value/s given.
Stardard deviation is given by\[\sigma=\sqrt{npq}\]. Similarly, variance is the square of standard deviation.
Given the question, they want the proability between 73 and 75, that is\[73<x <75\]
The formula will than be in form of Z, as the normal distribution table has vales in this variable, which is given by:
\[P=\frac{ X _{1}-\mu }{ \sigma } <Z <\frac{ X _{2}-\mu }{ \sigma } \]
\[P=\frac{ 73-74 }{ 5 }<z <\frac{ 75-74 }{ 5 }=-0.2<z <0.2=\left| z \right|<0.2\]

Answer 8

The probability of a modulus in the form given above is calculated by \[2P(Z <2)-1\]
To find the probability of Z, use the normal distribution table.
The value for Z being les than 2 is 0.9772.
The final result is then\[2(0.9772)-1=0.9544\]