Given a normal distribution with u = 75 and o = 40, if you select a sample of n = 16, a) what is the probability that the sample mean is above 95? (4 d. p.) b) What is the value, of which there is 97.5% chance that a sample mean is less than that value? (1 d. p.)

a) 0.0228

b) 94.6

Step-by-step explanation:

The formula for calculating a z-score when you are given a random.number of samples is z = (x-μ)/σ/√n

where x is the raw score

μ is the population mean

σ is the population standard deviation

n = random number of samples

Given a normal distribution with u = 75 and o = 40, if you select a sample of n = 16,

a) what is the probability that the sample mean is above 95? (4 d.p.) b)

= x = 95

Hence:

z = 95 - 75/40/√16

= 20/40/4

= 20/10

= 2

Probability value from Z-Table:

P(x<95) = 0.97725

P(x>95) = 1 - P(x<95) = 0.02275

The probability that the sample mean is above 95 to 4 decimal places = 0.0228

b) What is the value, of which there is 97.5% chance that a sample mean is less than that value?

97.5% chance = z score for the confidence interval = 1.96

Hence:

z = (x-μ)/σ/√n

1.96 = x - 75/40/√16

1.96 = x - 75/ 40/4

1.96 = x - 75/10

Cross Multiply

1.96 × 10 = x - 75

19.6 = x - 75

x = 19.6 + 75

x = 94.6

The value, of which there is 97.5% chance that a sample mean is less than that value is 94.6

coin flip

non-answer:

choosing

sorry if this isnt what you're looking for i dont really understand the question