In a normal distribution, the median is the same as the mean (25.3). The first quartile is the value of

such that

You have

For the standard normal distribution, the first quartile is about

, and by symmetry the third quartile would be

. In terms of the MCAT score distribution, these values are


The interquartile range (IQR) is just the difference between the two quartiles, so the IQR is about 8.8.
The central 80% of the scores have z-scores

such that

That leaves 10% on either side of this range, which means

You have

Converting to MCAT scores,


So the interval that contains the central 80% is

(give or take).
Answer:
The t value for 99% CI for 21 df is 2.831.
The critical value that should be used in constructing the confidence interval is (64.593, 86.407).
Step-by-step explanation:
Now the sample size is less than 30 and also population standard deviation is not known.
Then we will use t distribution to find CI
t value for 99% CI for 21 df is TINV(0.01,21)=2.831
The margin of error is 
Hence CI is
Answer
23/51
Step-by-step explanation:
The probability of the next customer paying with a debit or credit card is =23/51