Questions 15 and 16 are related | |||||||||||
15 | Suppose the distribution of sardine lengths is approximately normal with a mean of 4 centimeters and a standard deviation of 0.55 centimeter. Sardines that are too long or too short for a sardine can are ground up for cat food. The range of sizes that can be used in a sardine can is from 3 to 5 centimeters. What fraction of all sardines are ground up for cat food? | ||||||||||
a | 0.0292 | ||||||||||
b | 0.0688 | ||||||||||
c | 0.1010 | ||||||||||
d | 0.1142 | ||||||||||
16 | In the previous question what is the middle interval that includes 90% of all sardines? | ||||||||||
a | 3.4 | 4.6 | |||||||||
b | 3.3 | 4.7 | |||||||||
c | 3.2 | 4.8 | |||||||||
d | 3.1 | 4.9 |
Solution:-
15) (b) The fraction of all sardines are ground up for cat food is 0.0688.
Mean = 4, S.D = 0.55
x1 = 3.0
x2 = 5.0
By applying normal distruibution:-
z1 = - 1.82
z2 = 1.82
P( - 1.82 > z > 1.82) = P(z < - 1.82) + P(z > 1.82)
P( - 1.82 > z > 1.82) = 0.034 + 0.034
P( - 1.82 > z > 1.82) = 0.0688
16) (d) The middle interval that includes 90% of all sardines is C.I = (3.1, 4.9).
C.I = 4 + 1.645 × 0.55
C.I = 4 + 0.90475
C.I = (3.095, 4.905)
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