Question

The amount of time to complete a physical activity in a PE class is approximately normally normally distributed with a mean of 39.7 seconds and a standard deviation of 5.3 seconds.

a) What is the probability that a randomly chosen student completes the activity in less than 36.2 seconds?

b) What is the probability that a randomly chosen student completes the activity in more than 46.4 seconds?

c) What proportion of students take between 36.5 and 43.8 seconds to complete the activity?

d) 95% of all students finish the activity in less than  seconds.

Answer 1

Solution :

Given that ,

a) P(x < 36.2 )

= P[(x - $\mu$ ) / $\sigma$ < (36.2 - 39.7) / 5.3]

= P(z < -0.66)

Using z table,

= 0.2546

b) P(x > 46.4) = 1 - p( x< 46.4)

=1- p P[(x - $\mu$) / $\sigma$ < (46.4 - 39.7) / 5.3]

=1- P(z < 1.26)

= 1 - 0.8962

= 0.1038

c) P(36.5 < x < 43.8) = P[(36.5 - 39.7)/ 5.3) < (x - $\mu$) /$\sigma$  < (43.8 - 39.7) / 5.3) ]

= P(-0.60 < z < 0.77)

= P(z < 0.77) - P(z < -0.60)

Using z table,

= 0.7794 - 0.2743

= 0.5051

d) Using standard normal table,

P(Z < z) = 95%

= P(Z < 1.645 ) = 0.95  

z = 1.645

Using z-score formula,

x = z * $\sigma$ + $\mu$

x = 1.645 * 5.3 + 39.7

x = 48.4 seconds.