Agovernment agency has specialists who analyze the frequencies of letters of the alphabet in an attempt to decipher intercepted messages. in standard english text, a particular letter is used at a rate of 5.7%. a. find the mean and standard deviation for the number of times this letter will be found on a typical page of 1600 characters. muequals 91.2 (do not round.) sigmaequals 9.3 (round to one decimal place as needed.) b. in an intercepted message, a page of 1600 characters is found to have the letter occurring 102 times. is this unusual? \

A) μ = 91.2; σ = 9.3

B) No

Step-by-step explanation:

To find the mean, μ, we use the formula

μ = np, where n is the sample size and p is the probability of success (percentage of times the letter is used).

This gives us

μ = 0.057(1600) = 91.2

To find the standard deviation, we use the formula

σ = √(np(1-p)

This gives us

σ = √(1600×0.057×(1-0.057))

= √(1600×0.057×0.943) = √86.0016 = 9.2737 ≈ 9.3≈

Any value that is more than two standard deviations from the mean is considered unusual.  The value 102 is

(102-91.2)/9.3 = 10.8/9.3 = 1.16 standard deviations from the mean.  This is not statistically unusual.

it would take 7/10 of an hour too fill the pool.

step-by-step explanation: