Question

The mean number of students who went to examinations is 450 students with a standard deviation of 75 students. The distribution of the number of students is normal.

1. What number of students who correspond to the last 15%.
2. Fin the probability that less than 430 students want on campus examinations.
3. Calculate the probability that more than 400 students wanted on campus examinations
4. Determine the probability that between 452 and 463 students wanted on campus examinations

PUT DIAGRAMS PLEASE

Answer 1

Solution :

i.

Using standard normal table,

P(Z < z) = 15%

P(Z < -1.04) = 0.15

z = -1.04

Using z-score formula,

x = z * +

x = -1.04 * 75 + 450 = 372

372 number of students who correspond to the last 15%.

ii.

P(x < 430) = P[(x - ) / < (430 - 450) / 75]

= P(z < -0.2667)

= 0.3949

Probability = 0.3949

iii.

P(x > 400) = 1 - P(x < 400)

= 1 - P[(x - ) / < (400 - 450) / 75)

= 1 - P(z < -0.6667)

= 1 - 0.2525

= 0.7575

Probability = 0.7575

iv.

P(452 < x < 463) = P[(452 - 450)/ 75) < (x - ) /  < (463 - 450) / 75) ]

= P(0.0267 < z < 0.1733)

= P(z < 0.1733) - P(z < 0.0267)

= 0.5688 - 0.5107

= 0.0581

Probability = 0.0581