A population of values has a normal distribution with mu =67.9 and sigma =44.8. You intend to draw a random sample of size n=17.
Find the probability that a single randomly selected value is greater than 94

1 Answers

Best Answer

Step 1

Introduction:

Define X as the random variable of interest here.

It is given that X has a normal distribution with population mean mu = 67.9, and population standard deviation sigma = 44.8.

A sample of size n = 17 is selected from the population.

Step 2

Calculation:

The probability that a single randomly selected value is greater than 94 is calculated below:

P(X>94)=1-(X <= 94)

=1-P ((X- mu)/ (sigma) <= (94- mu)/ (sigma))

[By standardizing]

=1-P(Z <= (94-67.9)/(44.8))

[Z= (X- mu)/ (sigma) is the standard normal variable.]

approx 1-P(Z <= 0.58)

[The z-score is rounded to two decimal places.]

=1-0.7190

[Using the standard normal distribution table]

=0.2810

Thus, the probability that a single randomly selected value is greater than 94 is 0.2810.

Introduction:

Define X as the random variable of interest here.

It is given that X has a normal distribution with population mean mu = 67.9, and population standard deviation sigma = 44.8.

A sample of size n = 17 is selected from the population.

Step 2

Calculation:

The probability that a single randomly selected value is greater than 94 is calculated below:

P(X>94)=1-(X <= 94)

=1-P ((X- mu)/ (sigma) <= (94- mu)/ (sigma))

[By standardizing]

=1-P(Z <= (94-67.9)/(44.8))

[Z= (X- mu)/ (sigma) is the standard normal variable.]

approx 1-P(Z <= 0.58)

[The z-score is rounded to two decimal places.]

=1-0.7190

[Using the standard normal distribution table]

=0.2810

Thus, the probability that a single randomly selected value is greater than 94 is 0.2810.