A binomial distribution has p = 0.60 and n = 40. a. What are the mean and standard deviation for this​ distribution? b. What is the probability of exactly 27 ​successes? c. What is the probability of fewer than 30 ​successes? d. What is the probability of more than 20 ​successes?

a) mean = 24, standard deviation = 3.098386

b) the probability of exactly 27 ​successes is 0.08265

c) the probability of fewer than 30 ​successes is 0.96478

d) the probability of more than 20 ​successes is 0.8702

Step-by-step explanation:

Given the data in the question,

p = 0.60

n = 40

a) mean and standard deviation;

mean = np = 40 × 0.6

mean = 24

we know that Variance = np( 1 - p )

so standard deviation S.D = √( np( 1 - p ) )

S.D = √( 40 × 0.6( 1 - 0.6 ) )

S.D = √( 24( 0.4) )

S.D = √9.6

S.D = 3.098386

standard deviation = 3.098386

b) the probability of exactly 27 ​successes

P(  x=27 ) = ⁴⁰C₂₇ × ( 0.6 )²⁷ × ( 0.4 )⁽⁴⁰⁻²⁷⁾

= ⁴⁰C₂₇ × ( 0.6 )²⁷ × ( 0.4 )¹³

= [40! / ( 27!( 40 - 27 )! )] × ( 0.6 )²⁷ × ( 0.4 )¹³

= 0.08265

Hence, the probability of exactly 27 ​successes is 0.08265

c)  the probability of fewer than 30 ​successes.

P( x < 30 ) = 1 - P( x ≥ 30 )

= 1 - 0.3522248

= 0.96478

Hence, the probability of fewer than 30 ​successes is 0.96478

d) the probability of more than 20 ​successes.

p( x > 20 ) = 1 - P( x ≤ 20 )

= 1 - 0.1297657

= 0.8702

Hence, the probability of more than 20 ​successes is 0.8702

the expression (1.26)2t reveals the approximate rate of increases in the number of plants semiannually.

hope this : )

step-by-step explanation:

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