Answered

As a general rule, the sampling distribution of the sample proportions can be approximated by a normal probability distribution whenever

a)np greater than or equal to 5
b) n(1-p) greater than or equal to 5
c) n greater than or equal to 30
d) both a and b are true
A continuous random variable is uniformly distributed between a and b. The probability density function between a and b is
a) zero.
b) (a - b).
c) (b - a).
d) 1/(b - a).

A population has a mean of 84 and a standard deviation of 12. A sample of 36 observations will be taken. The probability that the sample mean will be between 80.54 and 88.9 is

a) 0.0347.
b) 0.7200.
c) 0.9511.
d) None of the alternative answers is correct.

Answer :

kingofkings3032
C cause it’s the right answer

In probability, the sampling distribution of the sample proportions can be approximated by a normal probability distribution when D. Both a and b are true.

How to calculate the probability?

The sampling distribution of the sample proportions can be approximated by a normal probability distribution whenever np is greater than or equal to 5 and when n(1-p) is greater than or equal to 5.

When the continuous random variable is uniformly distributed between a and b, the probability density function between a and b is 1/(b - a).

The probability that the sample mean will be between 80.54 and 88.9 will be:

= P[Z = 88.9 - 84)/(12/✓36)] - P(Z = 80.54 - 84)/(12/✓36)]

= P(Z = 2.45) - P(Z = -1.73)

= 0.9929 - 0.0418

= 0.9511

Therefore, the probability is 0.9511.

Learn more about probability on:

https://brainly.com/question/24756209

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