Question

A population has a mean of 74 with a standard deviation of 9.8.

a.) what is the probability that one element of the population selected at random is between 70 and 91?

b.) what is the probability that a random sample of 36 from this population has a sample mean between 73 and 79?

Answer 1

Solution :

Given that ,

mean = = 74

standard deviation = = 9.8

a.)

P(70 < x < 91) = P[(70 - 74)/ 9.8) < (x - ) /  < (91 - 74) / 9.8) ]

= P(-0.41 < z < 1.73)

= P(z < 1.73) - P(z < -0.41)

= 0.9582 - 0.3409

= 0.6173

Probability = 0.6173

b.)

= / n = 9.8 / 36 = 1.6333

= P[(73 - 74) /1.6333 < ( - ) / < (79 - 74) / 1.6333)]

= P(-0.61 < Z < 3.06)

= P(Z < 3.06) - P(Z < -0.61)

= 0.9989 - 0.2709

= 0.7280

Probability = 0.7280