Question

According to the Normal model ​N(0.052​,0.027) describing mutual fund returns in the 1st quarter of​ 2013, determine what percentage of this group of funds you would expect to have the following returns. Complete parts​ (a) through​ (d) below.

​a) Over​ 6.8%?	​b) Between​ 0% and​ 7.6%?
​c) More than​ 1%?	​d) Less than​ 0%?

Answer 1

Solution :

Given that ,

mean = = 0.052

standard deviation = = 0.027

(a) 6.8% = 0.068

P(x > 6.8%) = 1 - P(x < 0.068)

= 1 - P((x - ) / < (0.068 - 0.052) / 0.027)

= 1 - P(z < 0.5926)

= 1 - 0.7233   

= 0.2767

P(x > 6.8%) = 0.2767

Answer = 27.67%

(b)

P(0 < x < 0.076) = P((0 - 0.052)/ 0.027) < (x - ) / < (0.076 - 0.052) / 0.027) )

= P(-1.926 < z < 0.889)

= P(z < 0.889) - P(z < -1.926)

= 0.813 - 0.0271 = 0.7859

Answer = 78.59%

c)

P(x > 0.01) = 1 - P(x < 0.01)

= 1 - P((x - ) / < (0.01 - 0.052) / 0.027)

= 1 - P(z < -1.56)

= 1 - 0.0594   

= 0.9406

P(x > 0.01) = 0.9406

Answer = 94.06%

d)

P(x < 0) = P((x - ) / < (0 - 0.052) / 0.027)

= P(z < -1.926)

Using standard normal table,

P(x < 0) = 0.0271

Answer = 2.71%