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Oksana_A [137]
3 years ago
6

We choose 500 numbers uniformly at random from the interval[1.5,4.8].(a) Approximate the probability of the event that less than

65 of the numbers startwith the digit one.(b) Approximate the probability of the event that more than 160 of the numbersstart with the digit 3.
Mathematics
1 answer:
Ierofanga [76]3 years ago
3 0

Answer:

a = 0.0801

b = 0.1910

Step-by-step explanation:

a) probability of the event that numbers start with the digit 1 =P (1.5 < X <2)

=(2 - 1.5) / (4.8 - 1.5)=0.1515

mean =np

= 500 * 0.1515

= 75.76

std deviation =√(np (1 - p)) = 8.02)

probability of the event that less than 65 of the numbers start with the digit 1 = P(X < 65)

= P(Z < (64.5 - 75.76) / 8.02)

=P(Z < -1.4041)

= 0.0801

b) probability of the event of the numbers start with the digit 3 =P(3<X<4) =(4 - 3) / (4.8 - 1.5)

=0.3030

Here mean =np

= 500 * 0.3030

= 151.52

std deviation =√(np(1-p)) =10.28

probability of the event that more than 160 of the numbers start with the digit 3 = P(X > 160)

= 1 - P (Z < (160.5 - 151.52) / 10.28)

= 1 - P (Z < 0.8743)

= 1 - 0.8090

= 0.1910

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