A random sample of 18 graduates of a certain secretarial school typed an average of 80.6 words per minute with a standard deviation of 7.2 words per minute. Assuming a normal distribution for the number of words typed per minute, compute the 95% prediction interval for the next observed number of words per minute typed by a graduate of the secretarial school.

( 77.27 , 83.93)

Therefore at 95% confidence/prediction interval is

= ( 77.27 , 83.93)

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

x+/-zr/√n

Given that;

Mean x = 80.6 words per minute

Standard deviation r = 7.2

Number of samples n = 18

Confidence interval = 95%

z(at 95% confidence) = 1.96

Substituting the values we have;

80.6+/-1.96(7.2/√18)

80.6+/-1.96(1.697056274847)

80.6 +/- 3.33

= ( 77.27 , 83.93)

Therefore at 95% confidence/prediction interval is

= ( 77.27 , 83.93)

30.5*10=350$

you have to multiply the number without the decimal point then add it later but if its tenn u must move the place value of the decimal point 1 to the right.

130.66 for edgedunity students