In a large school district, it is known that 25% of all students entering kindergarten are already reading. A simple random sample of 10 new kindergarteners is drawn. What is the probability that fewer than three of them are able to read?

A .2503

B .2816

C .5000

D .5256

E .7759

Any formulas or ways to remember when to use a certain formula would be great, seeing as how I forgot how to do this stuff!

Given Information:

Probability of success = p = 25% = 0.25

Number of trials = n = 10   

Required Information:

P(x < 3) = ?

P(x < 3) = 0.5256 (option D)

Step-by-step explanation:

The given problem can be solved using binomial distribution since:

The binomial distribution is given by

P(x) = ⁿCₓ pˣ (1 - p)ⁿ⁻ˣ

Where n is the number of trials, x is the variable of interest and p is the probability of success.

The variable of interest in this case is fewer than 3 students and the probability that fewer than three students are able to read is

P(x < 3) = P(x = 0) + P(x = 1) + P(x = 2)

For P(x = 0):

Here we have x = 0, n = 10 and p = 0.25

P(x = 0) = ¹⁰C₀(0.25⁰)(1 - 0.25)¹⁰⁻⁰

P(x = 0) = (1)(0.25⁰)(0.75)¹⁰

P(x = 0) = 0.0563

For P(x = 1):

Here we have x = 1, n = 10 and p = 0.25

P(x = 1) = ¹⁰C₁(0.25¹)(1 - 0.25)¹⁰⁻¹

P(x = 1) = (10)(0.25¹)(0.75)⁹

P(x = 1) = 0.1877

For P(x = 2):

Here we have x = 2, n = 10 and p = 0.25

P(x = 2) = ¹⁰C₂(0.25²)(1 - 0.25)¹⁰⁻²

P(x = 2) = (45)(0.25²)(0.75)⁸

P(x = 2) = 0.28157

Finally,

P(x < 3) = P(x = 0) + P(x = 1) + P(x = 2)

P(x < 3) = 0.0563 + 0.1877 + 0.28157

P(x < 3) = 0.52557

Rounding off yields

P(x < 3) = 0.5256

Therefore, the probability that fewer than three of the students are able to read is 0.5256

sorry, i can't understand your question. will later if you describe more clearly!

: )