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Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.69 and standard deviation 0.74. in USE SALT (a) If a random sample of 25 specimens is selected, what is the probability that the sample average sediment density is at most 3.00? Between 2.69 and 3.00? (Round your answers to four decimal places. at most 3.00 between 2.69 and 3.00 (b) How large a sample size would be required to ensure that the first probability in part (a) is at least 0.99? (Round your answer up to the nearest whole number.)

Answer :

contexto1024

The formula to convert the sample mean to a z score is, z = (x−μ)/σ/√n

a) Probability that the sample average sediment density is at most 3 is 0.9525 and probability between 2.69 and 3.00 is 0.4525..

b) Large Sample size is approx. 4 ..

We have given that,

Suppose the sentiment density of a randomly selected specimen from a certain region is normally distributed with

Sample mean , X-bar = 2.69

Sample size , n = 25

Standard deviations, σ = 0.74

a) we have to see what's the probability that the sample average setiment is at most 3 and then we're going to calculate between 2.69 and 3 point, so the firstly we have to calculate the standard error.

standard error = σ/√n = 0.74/√25

=> standard error = 0.148

the probability that x is between 2.69 and 3 in the probability of 2.69, i.e P (2.69<z <3)

P(at most 3 ) = P( X-bar <3) = P( X-bar - u /σ /√n<( 3 - 2.69) /0.74/5 )

=> P(z< 1.67) = 0.9525

P (2.69<z <3) = P( 2.69 - 2.69/0.74/5<z< 1.67)

=> P( <z< 1.67) = 0.4525

b) A large sample size is needed to make sure the first probability is 0.99.

If P(X-bar <3) = 0.99

=> P( 0<z<(3-2.69)/0.74/√n ) = 0.99

=> (3 - 2.69 )/0.74 /√n = 2.33

=> 0.31 √n /0.25 = 2.33

=> n = ( 2.33×0.25/0.31)² = 3.5 ~ 4

Hence, sample size is 4 and P( <z< 1.67) is 0.4525

To learn more about Normal probability distribution, refer :

https://brainly.com/question/6476990

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