Time spent using e-mail per session is normally distributed, with mu equals 11 minutes and sigma equals 3 minutes. Assume that the time spent per session is normally distributed. Complete parts (a) through (d). a. If you select a random sample of 25 sessions, what is the probability that the sample mean is between 10.8 and 11.2 minutes? . 259 (Round to three decimal places as needed.) b. If you select a random sample of 25 sessions, what is the probability that the sample mean is between 10.5 and 11 minutes? . 297 (Round to three decimal places as needed.) c. If you select a random sample of 100 sessions, what is the probability that the sample mean is between 10.8 and 11.2 minutes? . 68 (Round to three decimal places as needed.)

a) 0.259

b) 0.297

c) 0.497

Step-by-step explanation:

To solve this problem, it is important to know the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean and standard deviation , a large sample size can be approximated to a normal distribution with mean and standard deviation

In this problem, we have that:

a. If you select a random sample of 25 sessions, what is the probability that the sample mean is between 10.8 and 11.2 minutes?

Here we have that

This probability is the pvalue of Z when X = 11.2 subtracted by the pvalue of Z when X = 10.8.

X = 11.2

By the Central Limit Theorem

has a pvalue of 0.6293.

X = 10.8

has a pvalue of 0.3707.

0.6293 - 0.3707 = 0.2586

0.259 probability, rounded to three decimal places.

b. If you select a random sample of 25 sessions, what is the probability that the sample mean is between 10.5 and 11 minutes?

Subtraction of the pvalue of Z when X = 11 subtracted by the pvalue of Z when X = 10.5. So

X = 11

has a pvalue of 0.5.

X = 10.5

has a pvalue of 0.2033.

0.5 - 0.2033 = 0.2967

0.297, rounded to three decimal places.

c. If you select a random sample of 100 sessions, what is the probability that the sample mean is between 10.8 and 11.2 minutes?

Here we have that

This probability is the pvalue of Z when X = 11.2 subtracted by the pvalue of Z when X = 10.8.

X = 11.2

By the Central Limit Theorem

has a pvalue of 0.7486.

X = 10.8

has a pvalue of 0.2514.

0.7486 - 0.2514 = 0.4972

0.497, rounded to three decimal places.

£10.1

Step-by-step explanation:

Highest exchange rate=1.3401

Lowest exchange rate=1.3199

The money Khan would have received more if he exchanged currencies at the highest rate compared to the lowest rate=500*(1.3401-1.3199)=£10.1