A restaurant chef ordered product to make a stew. He ordered 4 pounds of beef at 9.25 per pound, 6 pounds of potatoes at $6.75 a
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Answer:
98% confidence interval for the average age of all students is [24.302 , 25.698]
Step-by-step explanation:
We are given that a random sample of 36 students at a community college showed an average age of 25 years.
Also, assuming that the ages of all students at the college are normally distributed with a standard deviation of 1.8 years.
So, the pivotal quantity for 98% confidence interval for the average age is given by;
P.Q. = ~ N(0,1)
where, = sample average age = 25 years
= population standard deviation = 1.8 years
n = sample of students = 36
= population average age
So, 98% confidence interval for the average age, is ;
P(-2.3263 < N(0,1) < 2.3263) = 0.98
P(-2.3263 < < 2.3263) = 0.98
P( <
<
) = 0.98
P( <
<
) = 0.98
98% confidence interval for = [
,
]
= [ ,
]
= [24.302 , 25.698]
Therefore, 98% confidence interval for the average age of all students at this college is [24.302 , 25.698].