The probability computed shows that we can expect 1 out of 5 clothes to be clean.
From the information given, it was stated that 20% of the clothes are clean. Therefore, the probability of clean clothes will be:
= 20% = 1/5.
Therefore, one out fo five clothes are clean.
The standard deviation will be:
= ✓npq
where,
n = 5
p = 0.20
q = 1 - 0.20 = 0.80
SD = ✓(5 × 0.20 × 0.80)
= ✓0.80
= 0.09
The distribution is close to symmetric.
The binomial distribution table will be:
0 = (0.20)^0 (0.80)^5 = 0.3277
1 = (0.20)¹ (0.80)⁴ = 0.4096
2 = (0.20)² (0.80)³ = 0.2040
3 = (0.20)³ (0.80)² = 0.0512
4 = (0.20)⁴ (0.80)¹ = 0064
5 = (0.20)^5 (0.80)^0 = 0.0003
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